Find K-Length Substrings With No Repeated Characters in Python

Suppose we have a string S, we have to find the number of substrings of length K where no characters are repeated. So if S = “heyfriendshowareyou” and K is 5, then output will be 15, as the strings are [heyfr, eyfri, yfrie, frien, riend, iends, endsh, ndsho, dshow, showa, howar, oware, warey, areyo, reyou]

To solve this, we will follow these steps −

• create one empty map m, and left := 0 and right := -1 and ans := 0
• while right < length of string – 1
  • if right – left + 1 = k, then
    • increase ans by 1
    • decrease m[str[left]] by 1
    • increase left by 1
    • continue to next iteration
  • if str[right + 1] is not in m, then
    • set m[str[right + 1]] := 1
    • increase right by 1
  • else if m[str[right + 1]] is 0, then
    • increase m[str[right + 1]] by 1
    • increase right by 1
  • else
    • decrease m[str[left]] by 1
    • left := left + 1
• if right – left + 1 = k, then increase ans by 1
• return ans

Example

Let us see the following implementation to get a better understanding −

 Live Demo

class Solution(object):    def numKLenSubstrNoRepeats(self, S, K):       m = {}       left = 0       right = -1       ans = 0       while right<len(S)-1:          if right - left + 1 == K:             ans+=1             m[S[left]]-=1             left+=1             continue          if S[right+1] not in m :             m[S[right+1]]=1             right+=1          elif not m[S[right+1]]:             m[S[right+1]]+=1             right+=1          else:             m[S[left]]-=1             left+=1       if right - left + 1 == K:          ans+=1       return ans ob1 = Solution() print(ob1.numKLenSubstrNoRepeats("heyfriendshowareyou", 5))

Input

"heyfriendshowareyou" 5

Output

"AIIOC"