## Description

https://leetcode.com/problems/find-k-length-substrings-with-no-repeated-characters/

Given a string `s`

, return the number of substrings of length `k`

with no repeated characters.

**Example 1:**

Input:s = "havefunonleetcode", k = 5Output:6Explanation:There are 6 substrings they are : 'havef','avefu','vefun','efuno','etcod','tcode'.

**Example 2:**

Input:s = "home", k = 5Output:0Explanation:Notice k can be larger than the length of s. In this case is not possible to find any substring.

**Note:**

`1 <= s.length <= 10`

^{4}- All characters of s are lowercase English letters.
`1 <= k <= 10`

^{4}

## Explanation

Iterate each character and check if there is a string with no repeating character at length of k starting from the character.

## Python Solution

```
class Solution:
def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
if k > len(s):
return 0
count = 0
for i in range(len(s) - k + 1):
visited = set()
for j in range(i, i + k):
c = s[j]
if c in visited:
break
else:
visited.add(c)
if len(visited) == k:
count += 1
return count
```

- Time Complexity: O(N^2).
- Space Complexity: O(N).