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
- if right – left + 1 = k, then increase ans by 1
- return ans
Example
Let us see the following implementation to get a better understanding −
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"