Water collected between towers: Difference between revisions
m (→{{header|Haskell}}: Tidied) |
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<lang haskell>import Data.List (replicate, transpose) |
<lang haskell>import Data.List (replicate, transpose) |
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-------------- WATER COLLECTED BETWEEN TOWERS ------------ |
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towerPools :: [Int] -> [(Int, Int)] |
towerPools :: [Int] -> [(Int, Int)] |
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>>= zipWith ((<*>) (,) . (-)) |
>>= zipWith ((<*>) (,) . (-)) |
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--------------------------- TEST ------------------------- |
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main :: IO () |
main :: IO () |
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main = |
main = |
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[6, 7, 10, 7, 6] |
[6, 7, 10, 7, 6] |
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] |
] |
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------------------------- DIAGRAMS ----------------------- |
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display :: [(Int, Int)] -> String |
display :: [(Int, Int)] -> String |
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Revision as of 10:54, 18 May 2021
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
In a two-dimensional world, we begin with any bar-chart (or row of close-packed 'towers', each of unit width), and then it rains, completely filling all convex enclosures in the chart with water.
9 ██ 9 ██ 8 ██ 8 ██ 7 ██ ██ 7 ██≈≈≈≈≈≈≈≈██ 6 ██ ██ ██ 6 ██≈≈██≈≈≈≈██ 5 ██ ██ ██ ████ 5 ██≈≈██≈≈██≈≈████ 4 ██ ██ ████████ 4 ██≈≈██≈≈████████ 3 ██████ ████████ 3 ██████≈≈████████ 2 ████████████████ ██ 2 ████████████████≈≈██ 1 ████████████████████ 1 ████████████████████
In the example above, a bar chart representing the values [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] has filled, collecting 14 units of water.
Write a function, in your language, from a given array of heights, to the number of water units that can be held in this way, by a corresponding bar chart.
Calculate the number of water units that could be collected by bar charts representing each of the following seven series:
[[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]]
See, also:
- Four Solutions to a Trivial Problem – a Google Tech Talk by Guy Steele
- Water collected between towers on Stack Overflow, from which the example above is taken)
- An interesting Haskell solution, using the Tardis monad, by Phil Freeman in a Github gist.
11l
<lang 11l>F water_collected(tower)
V l = tower.len V highest_left = [0] [+] (1 .< l).map(n -> max(@tower[0 .< n])) V highest_right = (1 .< l).map(n -> max(@tower[n .< @l])) [+] [0] V water_level = (0 .< l).map(n -> max(min(@highest_left[n], @highest_right[n]) - @tower[n], 0)) print(‘highest_left: ’highest_left) print(‘highest_right: ’highest_right) print(‘water_level: ’water_level) print(‘tower_level: ’tower) print(‘total_water: ’sum(water_level)) print(‘’) R sum(water_level)
V towers = [
[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]]
print(towers.map(tower -> water_collected(tower)))</lang>
- Output:
highest_left: [0, 1, 5, 5, 7] highest_right: [7, 7, 7, 2, 0] water_level: [0, 0, 2, 0, 0] tower_level: [1, 5, 3, 7, 2] total_water: 2 highest_left: [0, 5, 5, 7, 7, 7, 7, 7, 9, 9] highest_right: [9, 9, 9, 9, 9, 9, 9, 2, 2, 0] water_level: [0, 2, 0, 5, 1, 3, 2, 0, 1, 0] tower_level: [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] total_water: 14 ... highest_left: [0, 6, 7, 10, 10] highest_right: [10, 10, 7, 6, 0] water_level: [0, 0, 0, 0, 0] tower_level: [6, 7, 10, 7, 6] total_water: 0 [2, 14, 35, 0, 0, 0, 0]
8080 Assembly
<lang 8080asm> org 100h jmp demo ;;; Calculate the amount of water a row of towers will hold ;;; Note: this will destroy the input array. ;;; Input: DE = tower array, BC = length of array ;;; Output: A = amount of water water: xra a ; Start with no water sta w_out+1 wscanr: mov h,d ; HL = right edge mov l,e dad b wscrlp: dcx h call cmp16 ; Reached beginning? jnc w_out ; Then stop mov a,m ; Otherwise, if current tower is zero ora a jz wscrlp ; Then keep scanning push b ; Keep length push d ; Keep array begin mvi b,0 ; No blocks yet xchg ; HL = left scanning edge, DE = right wscanl: mov a,m ; Get current column ora a ; Is zero? jz wunit ; Then see if an unit of water must be added dcr m ; Otherwise, decrease column inr b ; Increase blocks jmp wnext wunit: mov a,b ; Any blocks? ora a jz wnext lda w_out+1 ; If so, add water inr a sta w_out+1 wnext: inx h ; Next column call cmp16 jnc wscanl ; Until right edge reached mov a,b cmc ; Check if more than 1 block left rar ora a pop d ; Restore array begin pop b ; and length jnz wscanr ; If more than 1 block, keep scanning w_out: mvi a,0 ; Load water into A ret ;;; 16-bit compare DE to HL cmp16: mov a,d cmp h rnz mov a,e cmp l ret ;;; Calculate and print the amount of water for each input demo: lxi h,series load: mov e,m ; Load pointer inx h mov d,m inx h mov c,m ; Load length inx h mov b,m inx h mov a,d ; If pointer is zero, ora e rz ; stop. push h ; Otherwise, save the series pointer call water ; Calculate amount of water call printa ; Output amount of water pop h ; Restore series pointer jmp load ; Load next example ;;; Print A as integer value printa: lxi d,num ; Pointer to number string mvi c,10 ; Divisor digit: mvi b,-1 ; Quotient dloop: inr b ; Divide (by trial subtraction) sub c jnc dloop adi '0'+10 ; ASCII digit from remainder dcx d ; Store ASCII digit stax d mov a,b ; Continue with quotient ana a ; If not zero jnz digit mvi c,9 ; 9 = CP/M print string syscall jmp 5 ; Print number string db '***' ; Output number placeholder num: db ' $' ;;; Series t1: db 1,5,3,7,2 t2: db 5,3,7,2,6,4,5,9,1,2 t3: db 2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1 t4: db 5,5,5,5 t5: db 5,6,7,8 t6: db 8,7,7,6 t7: db 6,7,10,7,6 t_end: equ $ ;;; Lengths and pointers series: dw t1,t2-t1 dw t2,t3-t2 dw t3,t4-t3 dw t4,t5-t4 dw t5,t6-t5 dw t6,t7-t6 dw t7,t_end-t7 dw 0</lang>
- Output:
2 14 35 0 0 0 0
8086 Assembly
<lang asm> cpu 8086 org 100h section .text jmp demo ;;; Calculate the amount of water a row of towers will hold ;;; Note: this will destroy the input array. ;;; Input: DX = tower array, CX = length of array ;;; Output: AX = amount of water water: xor ax,ax ; Amount of water starts at zero xor bx,bx ; BH = zero, BL = block count .scanr: mov di,dx ; DI = right edge of towers add di,cx .rloop: dec di cmp di,dx ; Reached beginning? jl .out ; Then calculation is done. cmp bh,[di] ; Otherwise, if the tower is zero, je .rloop ; Keep scanning xor bl,bl ; Set block count to zero mov si,dx ; SI = left scanning edge .scanl: cmp bh,[si] ; Is the column empty? je .unit ; Then see whether to add an unit of water dec byte [si] ; Otherwise, remove block from tower inc bx ; And count it jmp .next .unit: test bl,bl ; Any blocks? jz .next inc ax ; If so, add unit of water .next: inc si ; Scan rightward cmp si,di ; Reached the right edge? jbe .scanl ; If not, keep going shr bl,1 ; If more than 1 block, jnz .scanr ; Keep going .out: ret ;;; Calculate and print the amount of water for each input demo: mov si,series .loop: lodsw ; Load pointer test ax,ax ; If 0, jz .done ; we're done. xchg ax,dx lodsw ; Load length xchg ax,cx push si ; Keep array pointer call water ; Calculate amount of water call prax ; Print AX pop si ; Restore array pointer jmp .loop .done: ret ;;; Print AX as number prax: mov bx,num ; Pointer to end of number string mov cx,10 ; Divisor .dgt: xor dx,dx ; Divide by 10 div cx add dl,'0' ; Add ASCII 0 to remainder dec bx ; Store digit mov [bx],dl test ax,ax ; If number not zero yet jnz .dgt ; Find rest of digits mov dx,bx ; Print number string mov ah,9 int 21h ret section .data db '*****' ; Output number placeholder num: db ' $' ;;; Series t1: db 1,5,3,7,2 t2: db 5,3,7,2,6,4,5,9,1,2 t3: db 2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1 t4: db 5,5,5,5 t5: db 5,6,7,8 t6: db 8,7,7,6 t7: db 6,7,10,7,6 t_end: equ $ ;;; Lengths and pointers series: dw t1,t2-t1 dw t2,t3-t2 dw t3,t4-t3 dw t4,t5-t4 dw t5,t6-t5 dw t6,t7-t6 dw t7,t_end-t7 dw 0</lang>
- Output:
2 14 35 0 0 0 0
AppleScript
<lang AppleScript>--------------- WATER COLLECTED BETWEEN TOWERS -------------
-- waterCollected :: [Int] -> Int on waterCollected(xs)
set leftWalls to scanl1(my max, xs)
set rightWalls to scanr1(my max, xs)
set waterLevels to zipWith(my min, leftWalls, rightWalls)
-- positive :: Num a => a -> Bool
script positive
on |λ|(x)
x > 0
end |λ|
end script
-- minus :: Num a => a -> a -> a
script minus
on |λ|(a, b)
a - b
end |λ|
end script
sum(filter(positive, zipWith(minus, waterLevels, xs)))
end waterCollected
TEST --------------------------
on run
map(waterCollected, ¬
[[1, 5, 3, 7, 2], ¬
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], ¬
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], ¬
[5, 5, 5, 5], ¬
[5, 6, 7, 8], ¬
[8, 7, 7, 6], ¬
[6, 7, 10, 7, 6]])
--> {2, 14, 35, 0, 0, 0, 0}
end run
GENERIC FUNCTIONS --------------------
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- init :: [a] -> [a] on init(xs)
if length of xs > 1 then
items 1 thru -2 of xs
else
{}
end if
end init
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- max :: Ord a => a -> a -> a on max(x, y)
if x > y then
x
else
y
end if
end max
-- min :: Ord a => a -> a -> a on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- scanl :: (b -> a -> b) -> b -> [a] -> [b] on scanl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl
-- scanl1 :: (a -> a -> a) -> [a] -> [a] on scanl1(f, xs)
if length of xs > 0 then
scanl(f, item 1 of xs, items 2 thru -1 of xs)
else
{}
end if
end scanl1
-- scanr :: (b -> a -> b) -> b -> [a] -> [b] on scanr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return reverse of lst
end tell
end scanr
-- scanr1 :: (a -> a -> a) -> [a] -> [a] on scanr1(f, xs)
if length of xs > 0 then
scanr(f, item -1 of xs, items 1 thru -2 of xs)
else
{}
end if
end scanr1
-- sum :: Num a => [a] -> a on sum(xs)
script add
on |λ|(a, b)
a + b
end |λ|
end script
foldl(add, 0, xs)
end sum
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] on zipWith(f, xs, ys)
set lng to min(length of xs, length of ys)
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, item i of ys)
end repeat
return lst
end tell
end zipWith</lang>
- Output:
<lang AppleScript>{2, 14, 35, 0, 0, 0, 0}</lang>
AWK
<lang AWK>
- syntax: GAWK -f WATER_COLLECTED_BETWEEN_TOWERS.AWK [-v debug={0|1}]
BEGIN {
wcbt("1,5,3,7,2")
wcbt("5,3,7,2,6,4,5,9,1,2")
wcbt("2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1")
wcbt("5,5,5,5")
wcbt("5,6,7,8")
wcbt("8,7,7,6")
wcbt("6,7,10,7,6")
exit(0)
} function wcbt(str, ans,hl,hr,i,n,tower) {
n = split(str,tower,",")
for (i=n; i>=0; i--) { # scan right to left
hr[i] = max(tower[i],(i<n)?hr[i+1]:0)
}
for (i=0; i<=n; i++) { # scan left to right
hl[i] = max(tower[i],(i!=0)?hl[i-1]:0)
ans += min(hl[i],hr[i]) - tower[i]
}
printf("%4d : %s\n",ans,str)
if (debug == 1) {
for (i=1; i<=n; i++) { printf("%-4s",tower[i]) } ; print("tower")
for (i=1; i<=n; i++) { printf("%-4s",hl[i]) } ; print("l-r")
for (i=1; i<=n; i++) { printf("%-4s",hr[i]) } ; print("r-l")
for (i=1; i<=n; i++) { printf("%-4s",min(hl[i],hr[i])) } ; print("min")
for (i=1; i<=n; i++) { printf("%-4s",min(hl[i],hr[i])-tower[i]) } ; print("sum\n")
}
} function max(x,y) { return((x > y) ? x : y) } function min(x,y) { return((x < y) ? x : y) } </lang>
- Output:
2 : 1,5,3,7,2 14 : 5,3,7,2,6,4,5,9,1,2 35 : 2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1 0 : 5,5,5,5 0 : 5,6,7,8 0 : 8,7,7,6 0 : 6,7,10,7,6
BASIC
<lang BASIC>10 DEFINT A-Z: DIM T(20): K=0 20 K=K+1: READ N: IF N=0 THEN END 30 FOR I=0 TO N-1: READ T(I): NEXT 40 W=0 50 FOR R=N-1 TO 0 STEP -1: IF T(R)=0 THEN NEXT ELSE IF R=0 THEN 110 60 B=0 70 FOR C=0 TO R 80 IF T(C)>0 THEN T(C)=T(C)-1: B=B+1 ELSE IF B>0 THEN W=W+1 90 NEXT 100 IF B>1 THEN 50 110 PRINT "Block";K;"holds";W;"water units." 120 GOTO 20 130 DATA 5, 1,5,3,7,2 140 DATA 10, 5,3,7,2,6,4,5,9,1,2 150 DATA 16, 2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1 160 DATA 4, 5,5,5,5 170 DATA 4, 5,6,7,8 180 DATA 4, 8,7,7,6 190 DATA 5, 6,7,10,7,6 200 DATA 0</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 holds 0 water units. Block 5 holds 0 water units. Block 6 holds 0 water units. Block 7 holds 0 water units.
C
Takes the integers as input from command line, prints out usage on incorrect invocation. <lang C>
- include<stdlib.h>
- include<stdio.h>
int getWater(int* arr,int start,int end,int cutoff){ int i, sum = 0;
for(i=start;i<=end;i++) sum += ((arr[cutoff] > arr[i])?(arr[cutoff] - arr[i]):0);
return sum; }
int netWater(int* arr,int size){ int i, j, ref1, ref2, marker, markerSet = 0,sum = 0;
if(size<3) return 0;
for(i=0;i<size-1;i++){ start:if(i!=size-2 && arr[i]>arr[i+1]){ ref1 = i;
for(j=ref1+1;j<size;j++){ if(arr[j]>=arr[ref1]){ ref2 = j;
sum += getWater(arr,ref1+1,ref2-1,ref1);
i = ref2;
goto start; }
else if(j!=size-1 && arr[j] < arr[j+1] && (markerSet==0||(arr[j+1]>=arr[marker]))){ marker = j+1; markerSet = 1; } }
if(markerSet==1){ sum += getWater(arr,ref1+1,marker-1,marker);
i = marker;
markerSet = 0;
goto start; } } }
return sum; }
int main(int argC,char* argV[]) { int *arr,i;
if(argC==1) printf("Usage : %s <followed by space separated series of integers>"); else{ arr = (int*)malloc((argC-1)*sizeof(int));
for(i=1;i<argC;i++) arr[i-1] = atoi(argV[i]);
printf("Water collected : %d",netWater(arr,argC-1)); }
return 0; } </lang> Output :
C:\rosettaCode>waterTowers.exe 1 5 3 7 2 Water collected : 2 C:\rosettaCode>waterTowers.exe 5 3 7 2 6 4 5 9 1 2 Water collected : 14 C:\rosettaCode>waterTowers.exe 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 Water collected : 35 C:\rosettaCode>waterTowers.exe 5 5 5 5 Water collected : 0 C:\rosettaCode>waterTowers.exe 8 7 7 6 Water collected : 0 C:\rosettaCode>waterTowers.exe 6 7 10 7 6 Water collected : 0
C#
Version 1
Translation from Visual Basic .NET. See that version 1 entry for code comment details and more sample output. <lang Csharp>class Program {
static void Main(string[] args)
{
int[][] wta = {
new int[] {1, 5, 3, 7, 2}, new int[] { 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
new int[] { 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
new int[] { 5, 5, 5, 5 }, new int[] { 5, 6, 7, 8 },
new int[] { 8, 7, 7, 6 }, new int[] { 6, 7, 10, 7, 6 }};
string blk, lf = "\n", tb = "██", wr = "≈≈", mt = " ";
for (int i = 0; i < wta.Length; i++)
{
int bpf; blk = ""; do
{
string floor = ""; bpf = 0; for (int j = 0; j < wta[i].Length; j++)
{
if (wta[i][j] > 0)
{ floor += tb; wta[i][j] -= 1; bpf += 1; }
else floor += (j > 0 && j < wta[i].Length - 1 ? wr : mt);
}
if (bpf > 0) blk = floor + lf + blk;
} while (bpf > 0);
while (blk.Contains(mt + wr)) blk = blk.Replace(mt + wr, mt + mt);
while (blk.Contains(wr + mt)) blk = blk.Replace(wr + mt, mt + mt);
if (args.Length > 0) System.Console.Write("\n{0}", blk);
System.Console.WriteLine("Block {0} retains {1,2} water units.",
i + 1, (blk.Length - blk.Replace(wr, "").Length) / 2);
}
}
}</lang>
- Output:
<lang>Block 1 retains 2 water units.
Block 2 retains 14 water units. Block 3 retains 35 water units. Block 4 retains 0 water units. Block 5 retains 0 water units. Block 6 retains 0 water units. Block 7 retains 0 water units.</lang>
Version 2
Conventional "scanning" algorithm, translated from the second version of Visual Basic.NET, but (intentionally tweaked to be) incapable of verbose output. See that version 2 entry for code comments and details. <lang cSharp>class Program { // Variable names key: // i Iterator (of the tower block array). // tba Tower block array. // tea Tower elevation array. // rht Right hand tower column number (position). // wu Water units (count). // bof Blocks on floor (count). // col Column number in elevation array (position).
static void Main(string[] args)
{
int i = 1; int[][] tba = {new int[] { 1, 5, 3, 7, 2 },
new int[] { 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
new int[] { 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
new int[] { 5, 5, 5, 5 }, new int[] { 5, 6, 7, 8 },
new int[] { 8, 7, 7, 6 }, new int[] { 6, 7, 10, 7, 6 }};
foreach (int[] tea in tba)
{
int rht, wu = 0, bof; do
{
for (rht = tea.Length - 1; rht >= 0; rht--)
if (tea[rht] > 0) break;
if (rht < 0) break;
bof = 0; for (int col = 0; col <= rht; col++)
{
if (tea[col] > 0) { tea[col] -= 1; bof += 1; }
else if (bof > 0) wu++;
}
if (bof < 2) break;
} while (true);
System.Console.WriteLine(string.Format("Block {0} {1} water units.",
i++, wu == 0 ? "does not hold any" : "holds " + wu.ToString()));
}
}
}</lang> Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.
C++
<lang cpp>
- include <iostream>
- include <vector>
- include <algorithm>
enum { EMPTY, WALL, WATER };
auto fill(const std::vector<int> b) {
auto water = 0;
const auto rows = *std::max_element(std::begin(b), std::end(b));
const auto cols = std::size(b);
std::vector<std::vector<int>> g(rows);
for (auto& r : g) {
for (auto i = 0; i < cols; ++i) {
r.push_back(EMPTY);
}
}
for (auto c = 0; c < cols; ++c) {
for (auto r = rows - 1u, i = 0u; i < b[c]; ++i, --r) {
g[r][c] = WALL;
}
}
for (auto c = 0; c < cols - 1; ++c) {
auto start_row = rows - b[c];
while (start_row < rows) {
if (g[start_row][c] == EMPTY) break;
auto c2 = c + 1;
bool hitWall = false;
while (c2 < cols) {
if (g[start_row][c2] == WALL) {
hitWall = true;
break;
}
++c2;
}
if (hitWall) {
for (auto i = c + 1; i < c2; ++i) {
g[start_row][i] = WATER;
++water;
}
}
++start_row;
}
}
return water;
}
int main() {
std::vector<std::vector<int>> b = {
{ 1, 5, 3, 7, 2 },
{ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
{ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
{ 5, 5, 5, 5 },
{ 5, 6, 7, 8 },
{ 8, 7, 7, 6 },
{ 6, 7, 10, 7, 6 }
};
for (const auto v : b) {
auto water = fill(v);
std::cout << water << " water drops." << std::endl;
}
std::cin.ignore();
std::cin.get();
return 0;
}</lang>
- Output:
2 water drops. 14 water drops. 35 water drops. 0 water drops. 0 water drops. 0 water drops. 0 water drops.
Clojure
Similar two passes algorithm as many solutions here. First traverse left to right to find the highest tower on the left of each position, inclusive of the tower at the current position, than do the same to find the highest tower to the right of each position. Finally, compute the total water units held at any position as the difference of those two heights.
<lang clojure> (defn trapped-water [towers]
(let [maxes #(reductions max %) ; the seq of increasing max values found in the input seq
maxl (maxes towers) ; the seq of max heights to the left of each tower
maxr (reverse (maxes (reverse towers))) ; the seq of max heights to the right of each tower
mins (map min maxl maxr)] ; minimum highest surrounding tower per position
(reduce + (map - mins towers)))) ; sum up the trapped water per position
</lang>
- Output:
<lang clojure>
- in the following, # is a tower block and ~ is trapped water
- 10|
- 9| #
- 8| #
- 7| # ~ ~ ~ ~ #
- 6| # ~ # ~ ~ #
- 5| # ~ # ~ # ~ # #
- 4| # ~ # ~ # # # #
- 3| # # # ~ # # # #
- 2| # # # # # # # # ~ #
- 1| # # # # # # # # # #
- ---+---------------------
- 5 3 7 2 6 4 5 9 1 2
(trapped-water [5 3 7 2 6 4 5 9 1 2]) ;; 14 </lang>
Cowgol
<lang cowgol>include "cowgol.coh"; include "argv.coh";
- Count the amount of water in a given array
sub water(towers: [uint8], length: intptr): (units: uint8) is
units := 0;
loop
var right := towers + length;
loop
right := @prev right;
if right < towers or [right] != 0 then
break;
end if;
end loop;
if right < towers then break; end if;
var blocks: uint8 := 0;
var col := towers;
while col <= right loop
if [col] != 0 then
[col] := [col] - 1;
blocks := blocks + 1;
elseif blocks != 0 then
units := units + 1;
end if;
col := @next col;
end loop;
if blocks < 2 then
break;
end if;
end loop;
end sub;
- Read list from the command line and print the answer
ArgvInit(); var towers: uint8[256]; var count: @indexof towers := 0; var n32: int32; loop
var argmt := ArgvNext();
if argmt == 0 as [uint8] then
break;
end if;
(n32, argmt) := AToI(argmt);
towers[count] := n32 as uint8;
count := count + 1;
end loop;
if count == 0 then
print("enter towers on command line\n");
ExitWithError();
end if;
print_i8(water(&towers[0], count as intptr)); print_nl();</lang>
- Output:
$ ./water.386 1 5 3 7 2 2 $ ./water.386 5 3 7 2 6 4 5 9 1 2 14 $ ./water.386 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 35 $ ./water.386 5 5 5 5 0 $ ./water.386 5 6 7 8 0 $ ./water.386 8 7 7 6 0 $ ./water.386 6 7 10 7 6 0
D
<lang D>import std.stdio;
void main() {
int i = 1;
int[][] tba = [
[ 1, 5, 3, 7, 2 ],
[ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ],
[ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ],
[ 5, 5, 5, 5 ],
[ 5, 6, 7, 8 ],
[ 8, 7, 7, 6 ],
[ 6, 7, 10, 7, 6 ]
];
foreach (tea; tba) {
int rht, wu, bof;
do {
for (rht = tea.length - 1; rht >= 0; rht--) {
if (tea[rht] > 0) {
break;
}
}
if (rht < 0) {
break;
}
bof = 0;
for (int col = 0; col <= rht; col++) {
if (tea[col] > 0) {
tea[col] -= 1; bof += 1;
} else if (bof > 0) {
wu++;
}
}
if (bof < 2) {
break;
}
} while (true);
write("Block ", i++);
if (wu == 0) {
write(" does not hold any");
} else {
write(" holds ", wu);
}
writeln(" water units.");
}
}</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.
Erlang
Implements a version that uses recursion to solve the problem functionally, using two passes without requiring list reversal or modifications. On the list iteration from head to tail, gather the largest element seen so far (being the highest one on the left). Once the list is scanned, each position returns the highest tower to its right as reported by its follower, along with the amount of water seen so far, which can then be used to calculate the value at the current position. Back at the first list element, the final result is gathered.
<lang erlang> -module(watertowers). -export([towers/1, demo/0]).
towers(List) -> element(2, tower(List, 0)).
tower([], _) -> {0,0}; tower([H|T], MaxLPrev) ->
MaxL = max(MaxLPrev, H),
{MaxR, WaterAcc} = tower(T, MaxL),
{max(MaxR,H), WaterAcc+max(0, min(MaxR,MaxL)-H)}.
demo() ->
Cases = [[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]],
[io:format("~p -> ~p~n", [Case, towers(Case)]) || Case <- Cases],
ok.
</lang>
- Output:
1> watertowers:demo(). [1,5,3,7,2] -> 2 [5,3,7,2,6,4,5,9,1,2] -> 14 [2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1] -> 35 [5,5,5,5] -> 0 [5,6,7,8] -> 0 [8,7,7,6] -> 0 [6,7,10,7,6] -> 0 ok
F#
see http://stackoverflow.com/questions/24414700/water-collected-between-towers/43779936#43779936 for an explanation of this code. It is proportional to the number of towers. Although the examples on stackoverflow claim this, the n they use is actually the distance between the two end towers and not the number of towers. Consider the case of a tower of height 5 at 1, a tower of height 10 at 39, and a tower of height 3 at 101. <lang fsharp> (* A solution I'd show to Euclid !!!!. Nigel Galloway May 4th., 2017
- )
let solve n =
let (n,_)::(i,e)::g = n|>List.sortBy(fun n->(-(snd n))) let rec fn i g e l = match e with | (n,e)::t when n < i -> fn n g t (l+(i-n-1)*e) | (n,e)::t when n > g -> fn i n t (l+(n-g-1)*e) | (n,t)::e -> fn i g e (l-t) | _ -> l fn (min n i) (max n i) g (e*(abs(n-i)-1))
</lang>
- Output:
solve [(1,1);(2,5);(3,3);(4,7);(5,2)] -> 2 solve [(1,5);(2,3);(3,7);(4,2);(5,6);(6,4);(7,5);(8,9);(9,1);(10,2)] -> 14 solve [(1,2);(2,6);(3,3);(4,5);(5,2);(6,8);(7,1);(8,4);(9,2);(10,2);(11,5);(12,3);(13,5);(14,7);(15,4);(16,1)] -> 35 solve [(1,5);(2,5);(3,5);(4,5)] -> 0 solve [(1,5);(2,6);(3,7);(4,8)] -> 0 solve [(1,8);(2,7);(3,7);(4,6)] -> 0 solve [(1,6);(2,7);(3,10);(4,7);(5,6)] -> 0 solve [(1,5);(39,10);(101,3)] -> 368
Factor
<lang factor>USING: formatting kernel math.statistics math.vectors sequences ;
- area ( seq -- n )
[ cum-max ] [ <reversed> cum-max reverse vmin ] [ v- sum ] tri ;
{
{ 1 5 3 7 2 }
{ 5 3 7 2 6 4 5 9 1 2 }
{ 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 }
{ 5 5 5 5 }
{ 5 6 7 8 }
{ 8 7 7 6 }
{ 6 7 10 7 6 }
} [ dup area "%[%d, %] -> %d\n" printf ] each</lang>
- Output:
{ 1, 5, 3, 7, 2 } -> 2
{ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 } -> 14
{ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 } -> 35
{ 5, 5, 5, 5 } -> 0
{ 5, 6, 7, 8 } -> 0
{ 8, 7, 7, 6 } -> 0
{ 6, 7, 10, 7, 6 } -> 0
FreeBASIC
Uses Nigel Galloway's very elegant idea, expressed verbosely so you can really see what's going on. <lang freebasic>type tower
hght as uinteger posi as uinteger
end type
sub shellsort( a() as tower )
'quick and dirty shellsort, not the focus of this exercise
dim as uinteger gap = ubound(a), i, j, n=ubound(a)
dim as tower temp
do
gap = int(gap / 2.2)
if gap=0 then gap=1
for i=gap to n
temp = a(i)
j=i
while j>=gap andalso a(j-gap).hght < temp.hght
a(j) = a(j - gap)
j -= gap
wend
a(j) = temp
next i
loop until gap = 1
end sub
'heights of towers in each city prefixed by the number of towers data 5, 1, 5, 3, 7, 2 data 10, 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 data 16, 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 data 4, 5, 5, 5, 5 data 4, 5, 6, 7, 8 data 4, 8, 7, 7, 6 data 5, 6, 7, 10, 7, 6
dim as uinteger i, n, j, first, last, water dim as tower manhattan(0 to 1) for i = 1 to 7
read n
redim manhattan( 0 to n-1 )
for j = 0 to n-1
read manhattan(j).hght
manhattan(j).posi = j
next j
shellsort( manhattan() )
if manhattan(0).posi < manhattan(1).posi then
first = manhattan(0).posi
last = manhattan(1).posi
else
first = manhattan(1).posi
last = manhattan(0).posi
end if
water = manhattan(1).hght * (last-first-1)
for j = 2 to n-1
if first<manhattan(j).posi and manhattan(j).posi<last then water -= manhattan(j).hght
if manhattan(j).posi < first then
water += manhattan(j).hght * (first-manhattan(j).posi-1)
first = manhattan(j).posi
end if
if manhattan(j).posi > last then
water += manhattan(j).hght * (manhattan(j).posi-last-1)
last = manhattan(j).posi
end if
next j
print using "City configuration ## collected #### units of water."; i; water
next i</lang>
- Output:
City configuration 1 collected 2 units of water. City configuration 2 collected 14 units of water. City configuration 3 collected 35 units of water. City configuration 4 collected 0 units of water. City configuration 5 collected 0 units of water. City configuration 6 collected 0 units of water. City configuration 7 collected 0 units of water.
Go
<lang go> package main
import "fmt"
func maxl(hm []int ) []int{ res := make([]int,len(hm)) max := 1 for i := 0; i < len(hm);i++{ if(hm[i] > max){ max = hm[i] } res[i] = max; } return res } func maxr(hm []int ) []int{ res := make([]int,len(hm)) max := 1 for i := len(hm) - 1 ; i >= 0;i--{ if(hm[i] > max){ max = hm[i] } res[i] = max; } return res } func min(a,b []int) []int { res := make([]int,len(a)) for i := 0; i < len(a);i++{ if a[i] >= b[i]{ res[i] = b[i] }else { res[i] = a[i] } } return res } func diff(hm, min []int) []int { res := make([]int,len(hm)) for i := 0; i < len(hm);i++{ if min[i] > hm[i]{ res[i] = min[i] - hm[i] } } return res } func sum(a []int) int { res := 0 for i := 0; i < len(a);i++{ res += a[i] } return res }
func waterCollected(hm []int) int { maxr := maxr(hm) maxl := maxl(hm) min := min(maxr,maxl) diff := diff(hm,min) sum := sum(diff) return sum }
func main() {
fmt.Println(waterCollected([]int{1, 5, 3, 7, 2}))
fmt.Println(waterCollected([]int{5, 3, 7, 2, 6, 4, 5, 9, 1, 2}))
fmt.Println(waterCollected([]int{2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1}))
fmt.Println(waterCollected([]int{5, 5, 5, 5}))
fmt.Println(waterCollected([]int{5, 6, 7, 8}))
fmt.Println(waterCollected([]int{8, 7, 7, 6}))
fmt.Println(waterCollected([]int{6, 7, 10, 7, 6}))
}</lang>
- Output:
2 14 35 0 0 0 0
Groovy
<lang Groovy> Integer waterBetweenTowers(List<Integer> towers) {
// iterate over the vertical axis. There the amount of water each row can hold is
// the number of empty spots, minus the empty spots at the beginning and end
return (1..towers.max()).collect { height ->
// create a string representing the row, '#' for tower material and ' ' for air
// use .trim() to remove spaces at beginning and end and then count remaining spaces
towers.collect({ it >= height ? "#" : " " }).join("").trim().count(" ")
}.sum()
}
tasks = [
[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]
]
tasks.each {
println "$it => total water: ${waterBetweenTowers it}"
} </lang>
- Output:
[1, 5, 3, 7, 2] => total water: 2 [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] => total water: 14 [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] => total water: 35 [5, 5, 5, 5] => total water: 0 [5, 6, 7, 8] => total water: 0 [8, 7, 7, 6] => total water: 0 [6, 7, 10, 7, 6] => total water: 0
Haskell
Following the approach of slightly modified cdk's Haskell solution at Stack Overflow. As recommended in Programming as if the Correct Data Structure (and Performance) Mattered it uses Vector instead of Array:
<lang haskell>import Data.Vector.Unboxed (Vector) import qualified Data.Vector.Unboxed as V
waterCollected :: Vector Int -> Int waterCollected =
V.sum . -- Sum of the water depths over each of V.filter (> 0) . -- the columns that are covered by some water. (V.zipWith (-) =<< -- Where coverages are differences between: (V.zipWith min . -- the lower water level in each case of: V.scanl1 max <*> -- highest wall to left, and V.scanr1 max)) -- highest wall to right.
main :: IO () main =
mapM_ (print . waterCollected . V.fromList) [ [1, 5, 3, 7, 2] , [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] , [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] , [5, 5, 5, 5] , [5, 6, 7, 8] , [8, 7, 7, 6] , [6, 7, 10, 7, 6] ]</lang>
- Output:
2 14 35 0 0 0 0
Or, using Data.List for simplicity - no need to prioritize performance here - and adding diagrams:
<lang haskell>import Data.List (replicate, transpose)
WATER COLLECTED BETWEEN TOWERS ------------
towerPools :: [Int] -> [(Int, Int)] towerPools =
zipWith min . scanl1 max <*> scanr1 max >>= zipWith ((<*>) (,) . (-))
TEST -------------------------
main :: IO () main =
mapM_
(putStrLn . display . towerPools)
[ [1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
]
DIAGRAMS -----------------------
display :: [(Int, Int)] -> String display = (<>) . showTowers <*> (('\n' :) . showLegend)
showTowers :: [(Int, Int)] -> String showTowers xs =
let upper = maximum (fst <$> xs)
in '\n' :
( unlines
. transpose
. fmap
( \(x, d) ->
concat $
replicate (upper - (x + d)) " "
<> replicate d "x"
<> replicate x "█"
)
)
xs
showLegend :: [(Int, Int)] -> String showLegend =
((<>) . show . fmap fst)
<*> ((" -> " <>) . show . foldr ((+) . snd) 0)</lang>
- Output:
█
█
█x█
█x█
███
████
█████
[1,5,3,7,2] -> 2
█
█
█xxxx█
█x█xx█
█x█x█x██
█x█x████
███x████
████████x█
██████████
[5,3,7,2,6,4,5,9,1,2] -> 14
█
█xxxxxxx█
█xxx█xxxxxxx█
█x█x█xxxx█x██
█x█x█x█xx█x███
███x█x█xx█████
██████x████████
████████████████
[2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1] -> 35
████
████
████
████
████
[5,5,5,5] -> 0
█
██
███
████
████
████
████
████
[5,6,7,8] -> 0
█
███
████
████
████
████
████
████
[8,7,7,6] -> 0
█
█
█
███
█████
█████
█████
█████
█████
█████
[6,7,10,7,6] -> 0
J
Inspired by #Julia.
Solution: <lang j>collectLevels =: >./\ <. >./\. NB. collect levels after filling waterLevels=: collectLevels - ] NB. water levels for each tower collectedWater=: +/@waterLevels NB. sum the units of water collected printTowers =: ' ' , [: |.@|: '#~' #~ ] ,. waterLevels NB. print a nice graph of towers and water</lang>
Examples: <lang j> collectedWater 5 3 7 2 6 4 5 9 1 2 14
printTowers 5 3 7 2 6 4 5 9 1 2
#
#
#~~~~#
#~#~~#
- ~#~#~##
- ~#~####
- ~####
- ~#
- ~#
- ~####
NB. Test cases
TestTowers =: <@".;._2 noun define
1 5 3 7 2 5 3 7 2 6 4 5 9 1 2 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 5 5 5 5 5 6 7 8 8 7 7 6 6 7 10 7 6 )
TestResults =: 2 14 35 0 0 0 0 TestResults -: collectedWater &> TestTowers NB. check tests
1</lang>
Java
<lang Java>public class WaterBetweenTowers {
public static void main(String[] args) {
int i = 1;
int[][] tba = new int[][]{
new int[]{1, 5, 3, 7, 2},
new int[]{5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
new int[]{2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
new int[]{5, 5, 5, 5},
new int[]{5, 6, 7, 8},
new int[]{8, 7, 7, 6},
new int[]{6, 7, 10, 7, 6}
};
for (int[] tea : tba) {
int rht, wu = 0, bof;
do {
for (rht = tea.length - 1; rht >= 0; rht--) {
if (tea[rht] > 0) {
break;
}
}
if (rht < 0) {
break;
}
bof = 0;
for (int col = 0; col <= rht; col++) {
if (tea[col] > 0) {
tea[col]--;
bof += 1;
} else if (bof > 0) {
wu++;
}
}
if (bof < 2) {
break;
}
} while (true);
System.out.printf("Block %d", i++);
if (wu == 0) {
System.out.print(" does not hold any");
} else {
System.out.printf(" holds %d", wu);
}
System.out.println(" water units.");
}
}
}</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.
JavaScript
ES5
<lang JavaScript>(function () {
'use strict';
// waterCollected :: [Int] -> Int
var waterCollected = function (xs) {
return sum( // water above each bar
zipWith(function (a, b) {
return a - b; // difference between water level and bar
},
zipWith(min, // lower of two flanking walls
scanl1(max, xs), // highest walls to left
scanr1(max, xs) // highest walls to right
),
xs // tops of bars
)
.filter(function (x) {
return x > 0; // only bars with water above them
})
);
};
// GENERIC FUNCTIONS ----------------------------------------
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
var zipWith = function (f, xs, ys) {
var ny = ys.length;
return (xs.length <= ny ? xs : xs.slice(0, ny))
.map(function (x, i) {
return f(x, ys[i]);
});
};
// scanl1 is a variant of scanl that has no starting value argument
// scanl1 :: (a -> a -> a) -> [a] -> [a]
var scanl1 = function (f, xs) {
return xs.length > 0 ? scanl(f, xs[0], xs.slice(1)) : [];
};
// scanr1 is a variant of scanr that has no starting value argument
// scanr1 :: (a -> a -> a) -> [a] -> [a]
var scanr1 = function (f, xs) {
return xs.length > 0 ? scanr(f, xs.slice(-1)[0], xs.slice(0, -1)) : [];
};
// scanl :: (b -> a -> b) -> b -> [a] -> [b]
var scanl = function (f, startValue, xs) {
var lst = [startValue];
return xs.reduce(function (a, x) {
var v = f(a, x);
return lst.push(v), v;
}, startValue), lst;
};
// scanr :: (b -> a -> b) -> b -> [a] -> [b]
var scanr = function (f, startValue, xs) {
var lst = [startValue];
return xs.reduceRight(function (a, x) {
var v = f(a, x);
return lst.push(v), v;
}, startValue), lst.reverse();
};
// sum :: (Num a) => [a] -> a
var sum = function (xs) {
return xs.reduce(function (a, x) {
return a + x;
}, 0);
};
// max :: Ord a => a -> a -> a
var max = function (a, b) {
return a > b ? a : b;
};
// min :: Ord a => a -> a -> a
var min = function (a, b) {
return b < a ? b : a;
};
// TEST ---------------------------------------------------
return [
[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
].map(waterCollected);
//--> [2, 14, 35, 0, 0, 0, 0]
})();</lang>
- Output:
<lang JavaScript>[2, 14, 35, 0, 0, 0, 0]</lang>
ES6
<lang JavaScript>(() => {
'use strict';
// ---------- WATER COLLECTED BETWEEN TOWERS -----------
// waterCollected :: [Int] -> Int
const waterCollected = xs =>
sum(filter(lt(0))(
zipWith(subtract)(xs)(
zipWith(min)(
scanl1(max)(xs)
)(
scanr1(max)(xs)
)
)
));
// ----------------------- TEST ------------------------
const main = () => [
[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
].map(waterCollected);
// ---------------------- GENERIC ----------------------
// Tuple (,) :: a -> b -> (a, b)
const Tuple = a =>
b => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// filter :: (a -> Bool) -> [a] -> [a]
const filter = p =>
// The elements of xs which match
// the predicate p.
xs => [...xs].filter(p);
// gt :: Ord a => a -> a -> Bool
const gt = x => y =>
'Tuple' === x.type ? (
x[0] > y[0]
) : (x > y);
// length :: [a] -> Int
const length = xs =>
// Returns Infinity over objects without finite
// length. This enables zip and zipWith to choose
// the shorter argument when one is non-finite,
// like cycle, repeat etc
'GeneratorFunction' !== xs.constructor
.constructor.name ? (
xs.length
) : Infinity;
// lt (<) :: Ord a => a -> a -> Bool
const lt = a =>
b => a < b;
// max :: Ord a => a -> a -> a
const max = a =>
// b if its greater than a,
// otherwise a.
b => a > b ? a : b;
// min :: Ord a => a -> a -> a
const min = a =>
b => b < a ? b : a;
// scanl :: (b -> a -> b) -> b -> [a] -> [b]
const scanl = f => startValue => xs =>
xs.reduce((a, x) => {
const v = f(a[0])(x);
return Tuple(v)(a[1].concat(v));
}, Tuple(startValue)([startValue]))[1];
// scanl1 :: (a -> a -> a) -> [a] -> [a]
const scanl1 = f =>
// scanl1 is a variant of scanl that has no
// starting value argument.
xs => xs.length > 0 ? (
scanl(f)(
xs[0]
)(xs.slice(1))
) : [];
// scanr :: (b -> a -> b) -> b -> [a] -> [b]
const scanr = f =>
startValue => xs => xs.reduceRight((a, x) => {
const v = f(x)(a[0]);
return Tuple(v)([v].concat(a[1]));
}, Tuple(startValue)([startValue]))[1];
// scanr1 :: (a -> a -> a) -> [a] -> [a]
const scanr1 = f =>
// scanr1 is a variant of scanr that has no
// seed-value argument, and assumes that
// xs is not empty.
xs => xs.length > 0 ? (
scanr(f)(
xs.slice(-1)[0]
)(xs.slice(0, -1))
) : [];
// subtract :: Num -> Num -> Num
const subtract = x =>
y => y - x;
// sum :: [Num] -> Num
const sum = xs =>
// The numeric sum of all values in xs.
xs.reduce((a, x) => a + x, 0);
// take :: Int -> [a] -> [a]
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => xs.slice(0, n);
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// A list constructed by zipping with a
// custom function, rather than with the
// default tuple constructor.
xs => ys => ((xs_, ys_) => {
const lng = Math.min(length(xs_), length(ys_));
return take(lng)(xs_).map(
(x, i) => f(x)(ys_[i])
);
})(xs, ys);
// MAIN --- return main();
})();</lang>
- Output:
<lang JavaScript>[2, 14, 35, 0, 0, 0, 0]</lang>
Julia
Inspired to #Python.
<lang julia>using Printf
function watercollected(towers::Vector{Int})
high_lft = vcat(0, accumulate(max, towers[1:end-1])) high_rgt = vcat(reverse(accumulate(max, towers[end:-1:2])), 0) waterlvl = max.(min.(high_lft, high_rgt) .- towers, 0) return waterlvl
end
function towerprint(towers, levels)
ctowers = copy(towers)
clevels = copy(levels)
hmax = maximum(towers)
ntow = length(towers)
for h in hmax:-1:1
@printf("%2i |", h)
for j in 1:ntow
if ctowers[j] + clevels[j] ≥ h
if clevels[j] > 0
cell = "≈≈"
clevels[j] -= 1
else
cell = "NN"
ctowers[j] -= 1
end
else
cell = " "
end
print(cell)
end
println("|")
end
println(" " * join(lpad(t, 2) for t in levels) * ": Water lvl")
println(" " * join(lpad(t, 2) for t in towers) * ": Tower lvl")
end
for towers in [[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]] towerprint(towers, watercollected(towers)) println()
end</lang>
- Output:
7 | NN |
6 | NN |
5 | NN≈≈NN |
4 | NN≈≈NN |
3 | NNNNNN |
2 | NNNNNNNN|
1 |NNNNNNNNNN|
0 0 2 0 0: Water lvl
1 5 3 7 2: Tower lvl
9 | NN |
8 | NN |
7 | NN≈≈≈≈≈≈≈≈NN |
6 | NN≈≈NN≈≈≈≈NN |
5 |NN≈≈NN≈≈NN≈≈NNNN |
4 |NN≈≈NN≈≈NNNNNNNN |
3 |NNNNNN≈≈NNNNNNNN |
2 |NNNNNNNNNNNNNNNN≈≈NN|
1 |NNNNNNNNNNNNNNNNNNNN|
0 2 0 5 1 3 2 0 1 0: Water lvl
5 3 7 2 6 4 5 9 1 2: Tower lvl
8 | NN |
7 | NN≈≈≈≈≈≈≈≈≈≈≈≈≈≈NN |
6 | NN≈≈≈≈≈≈NN≈≈≈≈≈≈≈≈≈≈≈≈≈≈NN |
5 | NN≈≈NN≈≈NN≈≈≈≈≈≈≈≈NN≈≈NNNN |
4 | NN≈≈NN≈≈NN≈≈NN≈≈≈≈NN≈≈NNNNNN |
3 | NNNNNN≈≈NN≈≈NN≈≈≈≈NNNNNNNNNN |
2 |NNNNNNNNNNNN≈≈NNNNNNNNNNNNNNNN |
1 |NNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN|
0 0 3 1 4 0 6 3 5 5 2 4 2 0 0 0: Water lvl
2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1: Tower lvl
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
5 5 5 5: Tower lvl
8 | NN|
7 | NNNN|
6 | NNNNNN|
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
5 6 7 8: Tower lvl
8 |NN |
7 |NNNNNN |
6 |NNNNNNNN|
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
8 7 7 6: Tower lvl
10 | NN |
9 | NN |
8 | NN |
7 | NNNNNN |
6 |NNNNNNNNNN|
5 |NNNNNNNNNN|
4 |NNNNNNNNNN|
3 |NNNNNNNNNN|
2 |NNNNNNNNNN|
1 |NNNNNNNNNN|
0 0 0 0 0: Water lvl
6 710 7 6: Tower lvl
Kotlin
<lang scala>// version 1.1.2
fun waterCollected(tower: IntArray): Int {
val n = tower.size
val highLeft = listOf(0) + (1 until n).map { tower.slice(0 until it).max()!! }
val highRight = (1 until n).map { tower.slice(it until n).max()!! } + 0
return (0 until n).map { maxOf(minOf(highLeft[it], highRight[it]) - tower[it], 0) }.sum()
}
fun main(args: Array<String>) {
val towers = listOf(
intArrayOf(1, 5, 3, 7, 2),
intArrayOf(5, 3, 7, 2, 6, 4, 5, 9, 1, 2),
intArrayOf(2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1),
intArrayOf(5, 5, 5, 5),
intArrayOf(5, 6, 7, 8),
intArrayOf(8, 7, 7, 6),
intArrayOf(6, 7, 10, 7, 6)
)
for (tower in towers) {
println("${"%2d".format(waterCollected(tower))} from ${tower.contentToString()}")
}
}</lang>
- Output:
2 from [1, 5, 3, 7, 2] 14 from [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] 35 from [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] 0 from [5, 5, 5, 5] 0 from [5, 6, 7, 8] 0 from [8, 7, 7, 6] 0 from [6, 7, 10, 7, 6]
Lua
<lang lua>function waterCollected(i,tower)
local length = 0
for _ in pairs(tower) do
length = length + 1
end
local wu = 0
repeat
local rht = length - 1
while rht >= 0 do
if tower[rht + 1] > 0 then
break
end
rht = rht - 1
end
if rht < 0 then
break
end
local bof = 0
local col = 0
while col <= rht do
if tower[col + 1] > 0 then
tower[col + 1] = tower[col + 1] - 1
bof = bof + 1
elseif bof > 0 then
wu = wu + 1
end
col = col + 1
end
if bof < 2 then
break
end
until false
if wu == 0 then
print(string.format("Block %d does not hold any water.", i))
else
print(string.format("Block %d holds %d water units.", i, wu))
end
end
function main()
local towers = {
{1, 5, 3, 7, 2},
{5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
{2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
{5, 5, 5, 5},
{5, 6, 7, 8},
{8, 7, 7, 6},
{6, 7, 10, 7, 6}
}
for i,tbl in pairs(towers) do
waterCollected(i,tbl)
end
end
main()</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water. Block 5 does not hold any water. Block 6 does not hold any water. Block 7 does not hold any water.
M2000 Interpreter
Scan min-max for each bar
<lang M2000 Interpreter> Module Water {
Flush ' empty stack
Data (1, 5, 3, 7, 2)
Data (5, 3, 7, 2, 6, 4, 5, 9, 1, 2)
Data (2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1)
Data (5, 5, 5, 5), (5, 6, 7, 8),(8, 7, 7, 6)
Data (6, 7, 10, 7, 6)
bars=stack.size ' mark stack frame
Dim bar()
for bar=1 to bars
bar()=Array ' pop an array from stack
acc=0
For i=1 to len(bar())-2
level1=bar(i)
level2=level1
m=each(bar(), i+1, 1)
while m
if array(m)>level1 then level1=array(m)
End While
n=each(bar(), i+1, -1)
while n
if array(n)>level2 then level2=array(n)
End While
acc+=max.data(min(level1, level2)-bar(i), 0)
Next i
Data acc ' push to end value
Next bar
finalwater=[] ' is a stack object
Print finalwater
} Water </lang>
Drain method
Module Water2 {
Flush ' empty stack
Data (1, 5, 3, 7, 2)
Data (5, 3, 7, 2, 6, 4, 5, 9, 1, 2)
Data (2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1)
Data (5, 5, 5, 5), (5, 6, 7, 8),(8, 7, 7, 6)
Data (6, 7, 10, 7, 6)
bars=stack.size ' mark stack frame
Dim bar()
For bar=1 to bars
bar()=Array ' pop an array from stack
acc=0
range=bar()#max()-bar()#min()
if range>0 then
dim water(len(bar()))=bar()#max()
water(0)=bar(0)
water(len(bar())-1)=bar(len(bar())-1)
For j=1 to range-1
For i=1 to len(bar())-2
if water(i)>bar(i) then if water(i-1)<water(i) Then water(i)--
Next i
For i=len(bar())-2 to 1
if water(i)>bar(i) then if water(i+1)<water(i) Then water(i)--
Next i
Next j
Data water()#sum()-bar()#sum()
Else
Data 0
End if
Next bar
finalwater=[]
Print finalwater
} Water2 <lang M2000 Interpreter> </lang>
Faster Method
<lang M2000 Interpreter> Module Water3 {
Flush ' empty stack
Data (1, 5, 3, 7, 2)
Data (5, 3, 7, 2, 6, 4, 5, 9, 1, 2)
Data (2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1)
Data (5, 5, 5, 5), (5, 6, 7, 8),(8, 7, 7, 6)
Data (6, 7, 10, 7, 6)
bars=stack.size ' mark stack frame
Dim bar()
for bar=1 to bars
bar()=Array ' pop an array from stack
acc=0
n=len(bar())-1
dim hl(n+1), hr(n+1)
For i=n to 0
hr(i)=max.data(bar(i), if(i<n->hr(i+1), 0))
Next i
For i=0 to n
hl(i)=max.data(bar(i), if(i>0->hl(i-1), 0))
acc+=min.data(hl(i), hr(i))-bar(i)
Next i
Data acc ' push to end value
Next bar
finalwater=[] ' is a stack object
Print finalwater
} Water3 </lang>
- Output:
2 14 35 0 0 0 0
Nim
<lang Nim>import math, sequtils, sugar
proc water(barChart: seq[int], isLeftPeak = false, isRightPeak = false): int =
if len(barChart) <= 2: return if isLeftPeak and isRightPeak: return sum(barChart[1..^2].map(x=>min(barChart[0], barChart[^1])-x)) var i: int if isLeftPeak: i = maxIndex(barChart[1..^1])+1 else: i = maxIndex(barChart[0..^2]) return water(barChart[0..i], isLeftPeak, true)+water(barChart[i..^1], true, isRightPeak)
const barCharts = [
@[1, 5, 3, 7, 2], @[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], @[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], @[5, 5, 5, 5], @[5, 6, 7, 8], @[8, 7, 7, 6], @[6, 7, 10, 7, 6]]
const waterUnits = barCharts.map(chart=>water(chart, false, false)) echo(waterUnits) </lang>
- Output:
@[2, 14, 35, 0, 0, 0, 0]
Perl
<lang perl>use Modern::Perl; use List::Util qw{ min max sum };
sub water_collected {
my @t = map { { TOWER => $_, LEFT => 0, RIGHT => 0, LEVEL => 0 } } @_;
my ( $l, $r ) = ( 0, 0 );
$_->{LEFT} = ( $l = max( $l, $_->{TOWER} ) ) for @t;
$_->{RIGHT} = ( $r = max( $r, $_->{TOWER} ) ) for reverse @t;
$_->{LEVEL} = min( $_->{LEFT}, $_->{RIGHT} ) for @t;
return sum map { $_->{LEVEL} > 0 ? $_->{LEVEL} - $_->{TOWER} : 0 } @t;
}
say join ' ', map { water_collected( @{$_} ) } (
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
);</lang>
- Output:
2 14 35 0 0 0 0
Phix
inefficient one-pass method
<lang Phix>function collect_water(sequence heights)
integer res = 0
for i=2 to length(heights)-1 do
integer lm = max(heights[1..i-1]),
rm = max(heights[i+1..$]),
d = min(lm,rm)-heights[i]
res += max(0,d)
end for
return res
end function
constant tests = {{1,5,3,7,2},
{5,3,7,2,6,4,5,9,1,2},
{2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1},
{5,5,5,5},
{5,6,7,8},
{8,7,7,6},
{6,7,10,7,6}}
for i=1 to length(tests) do
sequence ti = tests[i]
printf(1,"%35s : %d\n",{sprint(ti),collect_water(ti)})
end for</lang>
- Output:
{1,5,3,7,2} : 2
{5,3,7,2,6,4,5,9,1,2} : 14
{2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1} : 35
{5,5,5,5} : 0
{5,6,7,8} : 0
{8,7,7,6} : 0
{6,7,10,7,6} : 0
more efficient two-pass version
<lang Phix>function collect_water(sequence heights)
integer left_max = heights[1],
right_max = heights[$]
sequence left_height = heights,
right_height = heights
for i=2 to length(heights)-1 do
left_max = max(heights[i],left_max)
left_height[i] = left_max
right_max = max(heights[-i],right_max)
right_height[-i] = right_max
end for
sequence mins = sq_min(left_height,right_height),
diffs = sq_sub(mins,heights)
return sum(diffs)
end function</lang> (same output)
pretty print routine
<lang Phix>procedure print_water(sequence heights)
integer res = 0, l = length(heights)
sequence towers = repeat(repeat(' ',l),max(heights))
for i=1 to l do
for j=1 to heights[i] do
towers[-j][i] = '#'
end for
if i>1 and i<l then
integer lm = max(heights[1..i-1]),
rm = max(heights[i+1..$]),
m = min(lm,rm)
for j=heights[i]+1 to m do
towers[-j][i] = '~'
res += 1
end for
end if
end for
printf(1,"%s\ncollected:%d\n",{join(towers,"\n"),res})
end procedure
print_water({5,3,7,2,6,4,5,9,1,2})</lang>
- Output:
#
#
#~~~~#
#~#~~#
#~#~#~##
#~#~####
###~####
########~#
##########
collected:14
Phixmonti
<lang Phixmonti>include ..\Utilitys.pmt
def collect_water
0 var res
len 1 - 2 swap 2 tolist
for
var i
1 i 1 - slice max >ps
len i - 1 + i swap slice max >ps
i get ps> ps> min swap -
0 max res + var res
endfor
drop
res
enddef
( ( 1 5 3 7 2 )
( 5 3 7 2 6 4 5 9 1 2 ) ( 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 ) ( 5 5 5 5 ) ( 5 6 7 8 ) ( 8 7 7 6 ) ( 6 7 10 7 6 ) )
len for
get dup print " : " print collect_water ?
endfor</lang>
- Output:
[1, 5, 3, 7, 2] : 2 [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] : 14 [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] : 35 [5, 5, 5, 5] : 0 [5, 6, 7, 8] : 0 [8, 7, 7, 6] : 0 [6, 7, 10, 7, 6] : 0 === Press any key to exit ===
PicoLisp
<lang PicoLisp>(de water (Lst)
(sum
'((A)
(cnt
nT
(clip (mapcar '((B) (>= B A)) Lst)) ) )
(range 1 (apply max Lst)) ) )
(println
(mapcar
water
(quote
(1 5 3 7 2)
(5 3 7 2 6 4 5 9 1 2)
(2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1)
(5 5 5 5)
(5 6 7 8)
(8 7 7 6)
(6 7 10 7 6) ) ) )</lang>
- Output:
(2 14 35 0 0 0 0)
Python
Based on the algorithm explained at Stack Overflow:
<lang python>def water_collected(tower):
N = len(tower)
highest_left = [0] + [max(tower[:n]) for n in range(1,N)]
highest_right = [max(tower[n:N]) for n in range(1,N)] + [0]
water_level = [max(min(highest_left[n], highest_right[n]) - tower[n], 0)
for n in range(N)]
print("highest_left: ", highest_left)
print("highest_right: ", highest_right)
print("water_level: ", water_level)
print("tower_level: ", tower)
print("total_water: ", sum(water_level))
print("")
return sum(water_level)
towers = [[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]]
[water_collected(tower) for tower in towers]</lang>
- Output:
highest_left: [0, 1, 5, 5, 7] highest_right: [7, 7, 7, 2, 0] water_level: [0, 0, 2, 0, 0] tower_level: [1, 5, 3, 7, 2] total_water: 2 highest_left: [0, 5, 5, 7, 7, 7, 7, 7, 9, 9] highest_right: [9, 9, 9, 9, 9, 9, 9, 2, 2, 0] water_level: [0, 2, 0, 5, 1, 3, 2, 0, 1, 0] tower_level: [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] total_water: 14 highest_left: [0, 2, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8] highest_right: [8, 8, 8, 8, 8, 7, 7, 7, 7, 7, 7, 7, 7, 4, 1, 0] water_level: [0, 0, 3, 1, 4, 0, 6, 3, 5, 5, 2, 4, 2, 0, 0, 0] tower_level: [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] total_water: 35 highest_left: [0, 5, 5, 5] highest_right: [5, 5, 5, 0] water_level: [0, 0, 0, 0] tower_level: [5, 5, 5, 5] total_water: 0 highest_left: [0, 5, 6, 7] highest_right: [8, 8, 8, 0] water_level: [0, 0, 0, 0] tower_level: [5, 6, 7, 8] total_water: 0 highest_left: [0, 8, 8, 8] highest_right: [7, 7, 6, 0] water_level: [0, 0, 0, 0] tower_level: [8, 7, 7, 6] total_water: 0 highest_left: [0, 6, 7, 10, 10] highest_right: [10, 10, 7, 6, 0] water_level: [0, 0, 0, 0, 0] tower_level: [6, 7, 10, 7, 6] total_water: 0 [2, 14, 35, 0, 0, 0, 0]
Or, expressed in terms of itertools.accumulate, and showing diagrams:
<lang python>Water collected between towers
from itertools import accumulate from functools import reduce from operator import add
- ---------------------- TOWER POOLS -----------------------
- towerPools :: [Int] -> [(Int, Int)]
def towerPools(towers):
Tower heights with water depths.
def towerAndWater(level, tower):
return tower, level - tower
waterlevels = map(
min,
accumulate(towers, max),
reversed(list(
accumulate(reversed(towers), max)
)),
)
return list(map(towerAndWater, waterlevels, towers))
- ------------------------ DIAGRAMS ------------------------
- showTowers :: [(Int, Int)] -> String
def showTowers(xs):
Diagrammatic representation. upper = max(xs, key=fst)[0]
def row(xd):
return ' ' * (upper - add(*xd)) + (
snd(xd) * 'x' + '██' * fst(xd)
)
return unlines([
.join(x) for x in zip(*map(row, xs))
])
- showLegend :: (Int, Int)] -> String
def showLegend(xs):
String display of tower heights and
total sum of trapped water units.
towers, depths = zip(*xs)
return showList(towers) + (
' -> ' + str(sum(depths))
)
- -------------------------- TEST --------------------------
- main :: IO ()
def main():
Water collected in various flooded bar charts.
def diagram(xs):
return showTowers(xs) + '\n\n' + (
showLegend(xs) + '\n\n'
)
print(unlines(
map(compose(diagram, towerPools), [
[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
])
))
- ------------------------ GENERIC -------------------------
- compose :: ((a -> a), ...) -> (a -> a)
def compose(*fs):
Composition, from right to left,
of a series of functions.
def go(f, g):
return lambda x: f(g(x))
return reduce(go, fs, lambda x: x)
- fst :: (a, b) -> a
def fst(tpl):
First member of a pair. return tpl[0]
- showList :: [a] -> String
def showList(xs):
Stringification of a list. return '[' + ','.join(str(x) for x in xs) + ']'
- snd :: (a, b) -> b
def snd(tpl):
Second member of a pair. return tpl[1]
- unlines :: [String] -> String
def unlines(xs):
A single string formed by the intercalation
of a list of strings with the newline character.
return '\n'.join(xs)
- MAIN ---
if __name__ == '__main__':
main()
</lang>
- Output:
█
█
█x█
█x█
███
████
█████
█████
[1,5,3,7,2] -> 2
█
█
█xxxx█
█x█xx█
█x█x█x██
█x█x████
███x████
████████x█
██████████
██████████
[5,3,7,2,6,4,5,9,1,2] -> 14
█
█xxxxxxx█
█xxx█xxxxxxx█
█x█x█xxxx█x██
█x█x█x█xx█x███
███x█x█xx█████
██████x████████
████████████████
████████████████
[2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1] -> 35
████
████
████
████
████
████
████
████
████
████
[5,5,5,5] -> 0
█
██
███
████
████
████
████
████
████
████
████
████
████
[5,6,7,8] -> 0
█
███
████
████
████
████
████
████
████
████
████
████
████
████
[8,7,7,6] -> 0
█
█
█
███
█████
█████
█████
█████
█████
█████
█████
█████
█████
█████
█████
█████
[6,7,10,7,6] -> 0
Racket
<lang racket>#lang racket/base (require racket/match)
(define (water-collected-between-towers towers)
(define (build-tallest-left/rev-list t mx/l rv)
(match t
[(list) rv]
[(cons a d)
(define new-mx/l (max a mx/l))
(build-tallest-left/rev-list d new-mx/l (cons mx/l rv))]))
(define (collect-from-right t tallest/l mx/r rv)
(match t
[(list) rv]
[(cons a d)
(define new-mx/r (max a mx/r))
(define new-rv (+ rv (max (- (min new-mx/r (car tallest/l)) a) 0)))
(collect-from-right d (cdr tallest/l) new-mx/r new-rv)]))
(define reversed-left-list (build-tallest-left/rev-list towers 0 null)) (collect-from-right (reverse towers) reversed-left-list 0 0))
(module+ test
(require rackunit)
(check-equal?
(let ((towerss
'[[1 5 3 7 2]
[5 3 7 2 6 4 5 9 1 2]
[2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1]
[5 5 5 5]
[5 6 7 8]
[8 7 7 6]
[6 7 10 7 6]]))
(map water-collected-between-towers towerss))
(list 2 14 35 0 0 0 0)))</lang>
When run produces no output -- meaning that the tests have run successfully.
Raku
(formerly Perl 6)
<lang perl6>sub max_l ( @a ) { [\max] @a } sub max_r ( @a ) { ([\max] @a.reverse).reverse }
sub water_collected ( @towers ) {
return 0 if @towers <= 2;
my @levels = max_l(@towers) »min« max_r(@towers);
return ( @levels »-« @towers ).grep( * > 0 ).sum;
}
say map &water_collected,
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
- </lang>
- Output:
(2 14 35 0 0 0 0)
REXX
version 1
<lang rexx>/* REXX */ Call bars '1 5 3 7 2' Call bars '5 3 7 2 6 4 5 9 1 2' Call bars '2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1' Call bars '5 5 5 5' Call bars '5 6 7 8' Call bars '8 7 7 6' Call bars '6 7 10 7 6' Exit bars: Parse Arg bars bar.0=words(bars) high=0 box.=' ' Do i=1 To words(bars)
bar.i=word(bars,i) high=max(high,bar.i) Do j=1 To bar.i box.i.j='x' End End
m=1 w=0 Do Forever
Do i=m+1 To bar.0
If bar.i>bar.m Then
Leave
End
If i>bar.0 Then Leave
n=i
Do i=m+1 To n-1
w=w+bar.m-bar.i
Do j=bar.i+1 To bar.m
box.i.j='*'
End
End
m=n
End
m=bar.0 Do Forever
Do i=bar.0 To 1 By -1
If bar.i>bar.m Then
Leave
End
If i<1 Then Leave
n=i
Do i=m-1 To n+1 By -1
w=w+bar.m-bar.i
Do j=bar.i+1 To bar.m
box.i.j='*'
End
End
m=n
End
Say bars '->' w Call show Return show: Do j=high To 1 By -1
ol= Do i=1 To bar.0 ol=ol box.i.j End Say ol End
Return</lang>
- Output:
1 5 3 7 2 -> 2
x
x
x * x
x * x
x x x
x x x x
x x x x x
5 3 7 2 6 4 5 9 1 2 -> 14
x
x
x * * * * x
x * x * * x
x * x * x * x x
x * x * x x x x
x x x * x x x x
x x x x x x x x * x
x x x x x x x x x x
2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 -> 35
x
x * * * * * * * x
x * * * x * * * * * * * x
x * x * x * * * * x * x x
x * x * x * x * * x * x x x
x x x * x * x * * x x x x x
x x x x x x * x x x x x x x x
x x x x x x x x x x x x x x x x
5 5 5 5 -> 0
x x x x
x x x x
x x x x
x x x x
x x x x
5 6 7 8 -> 0
x
x x
x x x
x x x x
x x x x
x x x x
x x x x
x x x x
8 7 7 6 -> 0
x
x x x
x x x x
x x x x
x x x x
x x x x
x x x x
x x x x
6 7 10 7 6 -> 0
x
x
x
x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
version 2, simple numeric list output
<lang rexx>/*REXX program calculates and displays the amount of rainwater collected between towers.*/
call tower 1 5 3 7 2
call tower 5 3 7 2 6 4 5 9 1 2
call tower 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1
call tower 5 5 5 5
call tower 5 6 7 8
call tower 8 7 7 6
call tower 6 7 10 7 6
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ tower: procedure; arg y; #=words(y); t.=0; L.=0 /*the T. array holds the tower heights.*/
do j=1 for #; t.j= word(y, j) /*construct the towers, */
_= j-1; L.j= max(t._, L._) /* " " left─most tallest tower*/
end /*j*/
R.=0
do b=# by -1 for #; _= b+1; R.b= max(t._, R._) /*right─most tallest tower*/
end /*b*/
w.=0 /*rainwater collected.*/
do f=1 for #; if t.f>=L.f | t.f>=R.f then iterate /*rain between towers?*/
w.f= min(L.f, R.f) - t.f; w.00= w.00 + w.f /*rainwater collected.*/
end /*f*/
say right(w.00, 9) 'units of rainwater collected for: ' y /*display water units.*/
return</lang>
- output
2 units of rainwater collected for: 1 5 3 7 2
14 units of rainwater collected for: 5 3 7 2 6 4 5 9 1 2
35 units of rainwater collected for: 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1
0 units of rainwater collected for: 5 5 5 5
0 units of rainwater collected for: 5 6 7 8
0 units of rainwater collected for: 8 7 7 6
0 units of rainwater collected for: 6 7 10 7 6
version 3, with ASCII art
This REXX version shows a scale (showing the number of floors in the building) and a representation of the towers and water collected.
It tries to protect the aspect ratio by showing the buildings as in this task's preamble. <lang rexx>/*REXX program calculates and displays the amount of rainwater collected between towers.*/
call tower 1 5 3 7 2
call tower 5 3 7 2 6 4 5 9 1 2
call tower 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1
call tower 5 5 5 5
call tower 5 6 7 8
call tower 8 7 7 6
call tower 6 7 10 7 6
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ tower: procedure; arg y; #= words(y); t.=0; L.=0 /*the T. array holds the tower heights.*/
do j=1 for #; t.j=word(y,j); _=j-1 /*construct the towers; max height. */
L.j=max(t._, L._); t.0=max(t.0, t.j) /*left-most tallest tower; build scale.*/
end /*j*/
R.=0
do b=# by -1 for #; _= b+1; R.b= max(t._, R._) /*right-most tallest tower*/
end /*b*/
w.=0 /*rainwater collected.*/
do f=1 for #; if t.f>=L.f | t.f>=R.f then iterate /*rain between towers?*/
w.f= min(L.f, R.f) - t.f; w.00= w.00 + w.f /*rainwater collected.*/
end /*f*/
if w.00==0 then w.00= 'no' /*pretty up wording for "no rainwater".*/
ratio= 2 /*used to maintain a good aspect ratio.*/
p.= /*P. stores plot versions of towers. */
do c=0 to #; cc= c * ratio /*construct the plot+scale for display.*/
do h=1 for t.c+w.c; glyph= '█' /*maybe show a floor of some tower(s). */
if h>t.c then glyph= '≈' /* " " rainwater between towers. */
if c==0 then p.h= overlay(right(h, 9) , p.h, 1 ) /*tower scale*/
else p.h= overlay(copies(glyph,ratio) , p.h, 10+cc) /*build tower*/
end /*h*/
end /*c*/
p.1= overlay(w.00 'units of rainwater collected', p.1, 15*ratio+#) /*append text*/
do z=t.0 by -1 to 0; say p.z /*display various tower floors & water.*/
end /*z*/
return</lang>
- output
7 ██
6 ██
5 ██≈≈██
4 ██≈≈██
3 ██████
2 ████████
1 ██████████ 2 units of rainwater collected
9 ██
8 ██
7 ██≈≈≈≈≈≈≈≈██
6 ██≈≈██≈≈≈≈██
5 ██≈≈██≈≈██≈≈████
4 ██≈≈██≈≈████████
3 ██████≈≈████████
2 ████████████████≈≈██
1 ████████████████████ 14 units of rainwater collected
8 ██
7 ██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
6 ██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
5 ██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
4 ██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
3 ██████≈≈██≈≈██≈≈≈≈██████████
2 ████████████≈≈████████████████
1 ████████████████████████████████ 35 units of rainwater collected
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
8 ██
7 ████
6 ██████
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
8 ██
7 ██████
6 ████████
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
10 ██
9 ██
8 ██
7 ██████
6 ██████████
5 ██████████
4 ██████████
3 ██████████
2 ██████████
1 ██████████ no units of rainwater collected
Ruby
<lang ruby> def a(array) n=array.length left={} right={} left[0]=array[0] i=1 loop do
break if i >=n
left[i]=[left[i-1],array[i]].max
i += 1
end right[n-1]=array[n-1] i=n-2 loop do break if i<0
right[i]=[right[i+1],array[i]].max
i-=1 end i=0 water=0 loop do break if i>=n water+=[left[i],right[i]].min-array[i] i+=1 end puts water end
a([ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ]) a([ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ]) a([ 5, 5, 5, 5 ]) a([ 5, 6, 7, 8 ]) a([ 8, 7, 7, 6 ]) a([ 6, 7, 10, 7, 6 ]) return</lang> output
14 35 0 0 0 0
Rust
<lang rust> use std::cmp::min;
fn getfill(pattern: &[usize]) -> usize {
let mut total = 0;
for (idx, val) in pattern.iter().enumerate() {
let l_peak = pattern[..idx].iter().max();
let r_peak = pattern[idx + 1..].iter().max();
if l_peak.is_some() && r_peak.is_some() {
let peak = min(l_peak.unwrap(), r_peak.unwrap());
if peak > val {
total += peak - val;
}
}
}
total
}
fn main() {
let patterns = vec![
vec![1, 5, 3, 7, 2],
vec![5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
vec![2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
vec![5, 5, 5, 5],
vec![5, 6, 7, 8],
vec![8, 7, 7, 6],
vec![6, 7, 10, 7, 6],
];
for pattern in patterns {
println!("pattern: {:?}, fill: {}", &pattern, getfill(&pattern));
}
} </lang> output
pattern: [1, 5, 3, 7, 2], fill: 2 pattern: [5, 3, 7, 2, 6, 4, 5, 9, 1, 2], fill: 14 pattern: [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], fill: 35 pattern: [5, 5, 5, 5], fill: 0 pattern: [5, 6, 7, 8], fill: 0 pattern: [8, 7, 7, 6], fill: 0 pattern: [6, 7, 10, 7, 6], fill: 0
Scala
No sweat.
- Output:
See it yourself by running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>import scala.collection.parallel.CollectionConverters.VectorIsParallelizable
// Program to find maximum amount of water // that can be trapped within given set of bars. object TrappedWater extends App {
private val barLines = List( Vector(1, 5, 3, 7, 2), Vector(5, 3, 7, 2, 6, 4, 5, 9, 1, 2), Vector(2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1), Vector(5, 5, 5, 5), Vector(5, 6, 7, 8), Vector(8, 7, 7, 6), Vector(6, 7, 10, 7, 6)).zipWithIndex
// Method for maximum amount of water
private def sqBoxWater(barHeights: Vector[Int]): Int = {
def maxOfLeft = barHeights.par.scanLeft(0)(math.max).tail
def maxOfRight = barHeights.par.scanRight(0)(math.max).init
def waterlevels = maxOfLeft.zip(maxOfRight)
.map { case (maxL, maxR) => math.min(maxL, maxR) }
waterlevels.zip(barHeights).map { case (level, towerHeight) => level - towerHeight }.sum
}
barLines.foreach(barSet =>
println(s"Block ${barSet._2 + 1} could hold max. ${sqBoxWater(barSet._1)} units."))
}</lang>
Scheme
<lang scheme>(import (scheme base)
(scheme write))
(define (total-collected chart)
(define (highest-left vals curr)
(if (null? vals)
(list curr)
(cons curr
(highest-left (cdr vals) (max (car vals) curr)))))
(define (highest-right vals curr)
(reverse (highest-left (reverse vals) curr)))
;
(if (< (length chart) 3) ; catch the end cases
0
(apply +
(map (lambda (l c r)
(if (or (<= l c)
(<= r c))
0
(- (min l r) c)))
(highest-left chart 0)
chart
(highest-right chart 0)))))
(for-each
(lambda (chart) (display chart) (display " -> ") (display (total-collected chart)) (newline)) '((1 5 3 7 2) (5 3 7 2 6 4 5 9 1 2) (2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1) (5 5 5 5) (5 6 7 8) (8 7 7 6) (6 7 10 7 6)))</lang>
- Output:
(1 5 3 7 2) -> 2 (5 3 7 2 6 4 5 9 1 2) -> 14 (2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1) -> 35 (5 5 5 5) -> 0 (5 6 7 8) -> 0 (8 7 7 6) -> 0 (6 7 10 7 6) -> 0 (3 1 2) -> 1 (1) -> 0 () -> 0 (1 2) -> 0
Sidef
<lang ruby>func max_l(Array a, m = a[0]) {
gather { a.each {|e| take(m = max(m, e)) } }
}
func max_r(Array a) {
max_l(a.flip).flip
}
func water_collected(Array towers) {
var levels = (max_l(towers) »min« max_r(towers))
(levels »-« towers).grep{ _ > 0 }.sum
}
[
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
].map { water_collected(_) }.say</lang>
- Output:
[2, 14, 35, 0, 0, 0, 0]
Swift
<lang swift>// Based on this answer from Stack Overflow: // https://stackoverflow.com/a/42821623
func waterCollected(_ heights: [Int]) -> Int {
guard heights.count > 0 else {
return 0
}
var water = 0
var left = 0, right = heights.count - 1
var maxLeft = heights[left], maxRight = heights[right]
while left < right {
if heights[left] <= heights[right] {
maxLeft = max(heights[left], maxLeft)
water += maxLeft - heights[left]
left += 1
} else {
maxRight = max(heights[right], maxRight)
water += maxRight - heights[right]
right -= 1
}
}
return water
}
for heights in [[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]] {
print("water collected = \(waterCollected(heights))")
}</lang>
- Output:
water collected = 2 water collected = 14 water collected = 35 water collected = 0 water collected = 0 water collected = 0 water collected = 0
Tailspin
<lang tailspin> templates histogramWater
$ -> \( @: 0;
[$... -> { leftMax: $ -> #, value: $ } ] !
when <$@..> do @: $; $ !
otherwise $@ !
\) -> \( @: { rightMax: 0, sum: 0 };
$(last..1:-1)... -> #
$@.sum !
when <{ value: <$@.rightMax..> }> do @.rightMax: $.value;
when <{ value: <$.leftMax..> }> do !VOID
when <{ leftMax: <..$@.rightMax>}> do @.sum: $@.sum + $.leftMax - $.value;
otherwise @.sum: $@.sum + $@.rightMax - $.value;
\) !
end histogramWater
[[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]]... -> '$ -> histogramWater; water in $;$#10;' -> !OUT::write
</lang>
- Output:
2 water in [1, 5, 3, 7, 2] 14 water in [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] 35 water in [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] 0 water in [5, 5, 5, 5] 0 water in [5, 6, 7, 8] 0 water in [8, 7, 7, 6] 0 water in [6, 7, 10, 7, 6]
Tcl
Tcl makes for a surprisingly short and readable implementation, next to some of the more functional-oriented languages. <lang Tcl>namespace path {::tcl::mathfunc ::tcl::mathop}
proc flood {ground} {
set lefts [
set d 0
lmap g $ground {
set d [max $d $g]
}
]
set ground [lreverse $ground]
set rights [
set d 0
lmap g $ground {
set d [max $d $g]
}
]
set rights [lreverse $rights]
set ground [lreverse $ground]
set water [lmap l $lefts r $rights {min $l $r}]
set depths [lmap g $ground w $water {- $w $g}]
+ {*}$depths
}
foreach p {
{5 3 7 2 6 4 5 9 1 2}
{1 5 3 7 2}
{5 3 7 2 6 4 5 9 1 2}
{2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1}
{5 5 5 5}
{5 6 7 8}
{8 7 7 6}
{6 7 10 7 6}
} {
puts [flood $p]:\t$p
}</lang>
- Output:
14: 5 3 7 2 6 4 5 9 1 2 2: 1 5 3 7 2 14: 5 3 7 2 6 4 5 9 1 2 35: 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 0: 5 5 5 5 0: 5 6 7 8 0: 8 7 7 6 0: 6 7 10 7 6
Visual Basic .NET
Version 1
Method: Instead of "scanning" adjoining towers for each column, this routine converts the tower data into a string representation with building blocks, empty spaces, and potential water retention sites. The potential water retention sites are then "eroded" away where they are found to be unsupported. This is accomplished with the .Replace() function. The replace operations are unleashed upon the entire "block" of towers, rather than a cell at a time or a line at a time - which perhaps increases the program's execution-time, but reduces program's complexity.
The program can optionally display the interim string representation of each tower block before the final count is completed. I've since modified it to have the same block and wavy characters are the REXX 9.3 output, but used the double-wide columns, as pictured in the task definition area. <lang vbnet>' Convert tower block data into a string representation, then manipulate that. Module Module1
Sub Main(Args() As String)
Dim shoTow As Boolean = Environment.GetCommandLineArgs().Count > 1 ' Show towers.
Dim wta As Integer()() = { ' Water tower array (input data).
New Integer() {1, 5, 3, 7, 2}, New Integer() {5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
New Integer() {2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
New Integer() {5, 5, 5, 5}, New Integer() {5, 6, 7, 8},
New Integer() {8, 7, 7, 6}, New Integer() {6, 7, 10, 7, 6}}
Dim blk As String, ' String representation of a block of towers.
lf As String = vbLf, ' Line feed to separate floors in a block of towers.
tb = "██", wr = "≈≈", mt = " " ' Tower Block, Water Retained, eMpTy space.
For i As Integer = 0 To wta.Length - 1
Dim bpf As Integer ' Count of tower blocks found per floor.
blk = ""
Do
bpf = 0 : Dim floor As String = "" ' String representation of each floor.
For j As Integer = 0 To wta(i).Length - 1
If wta(i)(j) > 0 Then ' Tower block detected, add block to floor,
floor &= tb : wta(i)(j) -= 1 : bpf += 1 ' reduce tower by one.
Else ' Empty space detected, fill when not first or last column.
floor &= If(j > 0 AndAlso j < wta(i).Length - 1, wr, mt)
End If
Next
If bpf > 0 Then blk = floor & lf & blk ' Add floors until blocks are gone.
Loop Until bpf = 0 ' No tower blocks left, so terminate.
' Erode potential water retention cells from left and right.
While blk.Contains(mt & wr) : blk = blk.Replace(mt & wr, mt & mt) : End While
While blk.Contains(wr & mt) : blk = blk.Replace(wr & mt, mt & mt) : End While
' Optionaly show towers w/ water marks.
If shoTow Then Console.Write("{0}{1}", lf, blk)
' Subtract the amount of non-water mark characters from the total char amount.
Console.Write("Block {0} retains {1,2} water units.{2}", i + 1,
(blk.Length - blk.Replace(wr, "").Length) \ 2, lf)
Next
End Sub
End Module</lang>
- Output:
<lang>Block 1 retains 2 water units.
Block 2 retains 14 water units. Block 3 retains 35 water units. Block 4 retains 0 water units. Block 5 retains 0 water units. Block 6 retains 0 water units. Block 7 retains 0 water units.</lang> Verbose output shows towers with water ("Almost equal to" characters) left in the "wells" between towers. Just supply any command-line parameter to see it. Use no command line parameters to see the plain output above. <lang> ██
██ ██≈≈██ ██≈≈██ ██████ ████████
██████████ Block 1 retains 2 water units.
██
██
██≈≈≈≈≈≈≈≈██
██≈≈██≈≈≈≈██
██≈≈██≈≈██≈≈████ ██≈≈██≈≈████████ ██████≈≈████████ ████████████████≈≈██ ████████████████████ Block 2 retains 14 water units.
██
██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
██████≈≈██≈≈██≈≈≈≈██████████
████████████≈≈████████████████ ████████████████████████████████ Block 3 retains 35 water units.
████████ ████████ ████████ ████████ ████████ Block 4 retains 0 water units.
██ ████ ██████
████████ ████████ ████████ ████████ ████████ Block 5 retains 0 water units.
██ ██████ ████████ ████████ ████████ ████████ ████████ ████████ Block 6 retains 0 water units.
██ ██ ██ ██████
██████████ ██████████ ██████████ ██████████ ██████████ ██████████ Block 7 retains 0 water units.</lang>
Version 2
Method: More conventional "scanning" method. A Char array is used, but no Replace() statements. Output is similar to version 1, although there is now a left margin of three spaces, the results statement is immediately to the right of the string representation of the tower blocks (instead of underneath), the verb is "hold(s)" instead of "retains", and there is a special string when the results indicate zero.
<lang vbnet>Module Module1
<summary>
wide - Widens the aspect ratio of a linefeed separated string.
</summary>
<param name="src">A string representing a block of towers.</param>
<param name="margin">Optional padding for area to the left.</param>
<returns>A double-wide version of the string.</returns>
Function wide(src As String, Optional margin As String = "") As String
Dim res As String = margin : For Each ch As Char In src
res += If(ch < " ", ch & margin, ch + ch) : Next : Return res
End Function
<summary>
cntChar - Counts characters, also custom formats the output.
</summary>
<param name="src">The string to count characters in.</param>
<param name="ch">The character to be counted.</param>
<param name="verb">Verb to include in format. Expecting "hold",
but can work with "retain" or "have".</param>
<returns>The count of chars found in a string, and formats a verb.</returns>
Function cntChar(src As String, ch As Char, verb As String) As String
Dim cnt As Integer = 0
For Each c As Char In src : cnt += If(c = ch, 1, 0) : Next
Return If(cnt = 0, "does not " & verb & " any",
verb.Substring(0, If(verb = "have", 2, 4)) & "s " & cnt.ToString())
End Function
<summary>
report - Produces a report of the number of rain units found in
a block of towers, optionally showing the towers.
Autoincrements the blkID for each report.
</summary>
<param name="tea">An int array with tower elevations.</param>
<param name="blkID">An int of the block of towers ID.</param>
<param name="verb">The verb to use in the description.
Defaults to "has / have".</param>
<param name="showIt">When true, the report includes a string representation
of the block of towers.</param>
<returns>A string containing the amount of rain units, optionally preceeded by
a string representation of the towers holding any water.</returns>
Function report(tea As Integer(), ' Tower elevation array.
ByRef blkID As Integer, ' Block ID for the description.
Optional verb As String = "have", ' Verb to use in the description.
Optional showIt As Boolean = False) As String ' Show representaion.
Dim block As String = "", ' The block of towers.
lf As String = vbLf, ' The separator between floors.
rTwrPos As Integer ' The position of the rightmost tower of this floor.
Do
For rTwrPos = tea.Length - 1 To 0 Step -1 ' Determine the rightmost tower
If tea(rTwrPos) > 0 Then Exit For ' postition on this floor.
Next
If rTwrPos < 0 Then Exit Do ' When no towers remain, exit the do loop.
' init the floor to a space filled Char array, as wide as the block of towers.
Dim floor As Char() = New String(" ", tea.Length).ToCharArray()
Dim bpf As Integer = 0 ' The count of blocks found per floor.
For column As Integer = 0 To rTwrPos ' Scan from left to right.
If tea(column) > 0 Then ' If a tower exists here,
floor(column) = "█" ' mark the floor with a block,
tea(column) -= 1 ' drop the tower elevation by one,
bpf += 1 ' and advance the block count.
ElseIf bpf > 0 Then ' Otherwise, see if a tower is present to the left.
floor(column) = "≈" ' OK to fill with water.
End If
Next
If bpf > If(showIt, 0, 1) Then ' Continue the building only when needed.
' If not showing blocks, discontinue building when a single tower remains.
' build tower blocks string with each floor added to top.
block = New String(floor) & If(block = "", "", lf) & block
Else
Exit Do ' Ran out of towers, so exit the do loop.
End If
Loop While True ' Depending on previous break statements to terminate the do loop.
blkID += 1 ' increment block ID counter.
' format report and return it.
Return If(showIt, String.Format(vbLf & "{0}", wide(block, " ")), "") &
String.Format(" Block {0} {1} water units.", blkID, cntChar(block, "≈", verb))
End Function
<summary>
Main routine.
With one command line parameter, it shows tower blocks,
with no command line parameters, it shows a plain report
</summary>
Sub Main()
Dim shoTow As Boolean = Environment.GetCommandLineArgs().Count > 1 ' Show towers.
Dim blkCntr As Integer = 0 ' Block ID for reports.
Dim verb As String = "hold" ' "retain" or "have" can be used instead of "hold".
Dim tea As Integer()() = {New Integer() {1, 5, 3, 7, 2}, ' Tower elevation data.
New Integer() {5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
New Integer() {2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
New Integer() {5, 5, 5, 5}, New Integer() {5, 6, 7, 8},
New Integer() {8, 7, 7, 6}, New Integer() {6, 7, 10, 7, 6}}
For Each block As Integer() In tea
' Produce report for each block of towers.
Console.WriteLine(report(block, blkCntr, verb, shoTow))
Next
End Sub
End Module</lang> Regular version 2 output: <lang> Block 1 holds 2 water units.
Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.</lang>
Sample of version 2 verbose output: <lang> ██
██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
██████≈≈██≈≈██≈≈≈≈██████████
████████████≈≈████████████████
████████████████████████████████ Block 3 holds 35 water units.
████████ ████████ ████████ ████████ ████████ Block 4 does not hold any water units.</lang>
Wren
<lang ecmascript>import "/math" for Math, Nums import "/fmt" for Fmt
var waterCollected = Fn.new { |tower|
var n = tower.count
var highLeft = [0] + (1...n).map { |i| Nums.max(tower[0...i]) }.toList
var highRight = (1...n).map { |i| Nums.max(tower[i...n]) }.toList + [0]
var t = (0...n).map { |i| Math.max(Math.min(highLeft[i], highRight[i]) - tower[i], 0) }
return Nums.sum(t)
}
var towers = [
[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]
] for (tower in towers) Fmt.print("$2d from $n", waterCollected.call(tower), tower)</lang>
- Output:
2 from [1, 5, 3, 7, 2] 14 from [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] 35 from [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] 0 from [5, 5, 5, 5] 0 from [5, 6, 7, 8] 0 from [8, 7, 7, 6] 0 from [6, 7, 10, 7, 6]
Yabasic
<lang Yabasic>data 7 data "1,5,3,7,2", "5,3,7,2,6,4,5,9,1,2", "2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1" data "5,5,5,5", "5,6,7,8", "8,7,7,6", "6,7,10,7,6"
read n
for i = 1 to n read n$ wcbt(n$) next i
sub wcbt(s$) local tower$(1), hr(1), hl(1), n, i, ans, k
n = token(s$, tower$(), ",")
redim hr(n) redim hl(n) for i = n to 1 step -1 if i < n then k = hr(i + 1) else k = 0 end if hr(i) = max(val(tower$(i)), k) next i for i = 1 to n if i then k = hl(i - 1) else k = 0 end if hl(i) = max(val(tower$(i)), k) ans = ans + min(hl(i), hr(i)) - val(tower$(i)) next i print ans," ",n$ end sub</lang>
zkl
<lang zkl>fcn waterCollected(walls){
// compile max wall heights from left to right and right to left
// then each pair is left/right wall of that cell.
// Then the min of each wall pair == water height for that cell
scanl(walls,(0).max) // scan to right, f is max(0,a,b)
.zipWith((0).MAX.min, // f is MAX.min(a,b) == min(a,b)
scanl(walls.reverse(),(0).max).reverse()) // right to left
// now subtract the wall height from the water level and add 'em up
.zipWith('-,walls).filter('>(0)).sum(0);
} fcn scanl(xs,f,i=0){ // aka reduce but save list of results
xs.reduce('wrap(s,x,a){ s=f(s,x); a.append(s); s },i,ss:=List());
ss
} // scanl((1,5,3,7,2),max,0) --> (1,5,5,7,7)</lang> <lang zkl>T( T(1, 5, 3, 7, 2), T(5, 3, 7, 2, 6, 4, 5, 9, 1, 2),
T(2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1), T(5, 5, 5, 5), T(5, 6, 7, 8),T(8, 7, 7, 6), T(6, 7, 10, 7, 6) )
.pump(List, waterCollected).println();</lang>
- Output:
L(2,14,35,0,0,0,0)
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