Most frequent k chars distance: Difference between revisions
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;; in the given tests...)
</lang>
=={{header|Ring}}==
<lang ring>
# Project : Most frequent k chars distance
# Date : 2018/02/16
# Author : Gal Zsolt (~ CalmoSoft ~)
# Email : <calmosoft@gmail.com>
load "stdlib.ring"
str1 = "LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV"
str2 = "EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG"
//MostFreqKHashing(str1,2) = L9T8
//MostFreqKHashing(str2,2) = F9L8
//MostFreqKSDF(str1,str2,2,100) = 83
// A=65, Z=90
see "Str1 = " + str1 + nl
see "Str2 = " + str2 + nl
MostFreqKHashing(str1,"str1")
MostFreqKHashing(str2,"str2")
func MostFreqKHashing(str3,strp)
chr = newlist(26,2)
for n = 1 to 26
str = char(n+64)
cstr = count(str3,str)
chr[n][1] = str
chr[n][2] = cstr
next
chr = sortsecond(chr)
chr = reverse(chr)
see "MostFreqKHashing(" + strp + ",2) = "
see chr[1][1] + chr[1][2] + chr[2][1] + chr[2][2] + nl
func count(cString,dString)
sum = 0
while substr(cString,dString) > 0
sum++
cString = substr(cString,substr(cString,dString)+len(string(sum)))
end
return sum
func sortsecond(alist)
aList = sort(alist,2)
for n=1 to len(alist)-1
for m=n to len(aList)-1
if alist[m+1][2] = alist[m][2] and alist[m+1][1] < alist[m][1]
temp = alist[m+1]
alist[m+1] = alist[m]
alist[m] = temp ok
next
next
return aList
</lang>
Output:
<pre>
Str1 = LCLYTHIGRNIYYGSYLYSETWNTGIMLLLITMATAFMGYVLPWGQMSFWGATVITNLFSAIPYIGTNLV
Str2 = EWIWGGFSVDKATLNRFFAFHFILPFTMVALAGVHLTFLHETGSNNPLGLTSDSDKIPFHPYYTIKDFLG
MostFreqKHashing(str1,2) = L9T8
MostFreqKHashing(str2,2) = F9L8
</pre>
=={{header|Sidef}}==
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