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"