LeetCode 1100: Find K-Length Substrings With No Repeated Characters Solution

Master LeetCode problem 1100 (Find K-Length Substrings With No Repeated Characters), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1100. Find K-Length Substrings With No Repeated Characters

Medium

Hash Table String Sliding Window

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach where we maintain a window of size k and slide it through the given string. At each step, we check if the current window contains all unique characters. If it does, we increment the count of valid substrings by 1.

Here is the step-by-step algorithm:

Initialize a hashmap to store the count of characters in the current window.
Initialize variables left and right to represent the left and right pointers of the window.
Iterate through the string using the right pointer.
- Increment the count of the character at the right pointer.
- If the count of the character becomes greater than 1, increment the left pointer until the count becomes 1.
- Calculate the size of the current window and check if it is equal to k.
  - If it is equal to k, increment the count of valid substrings.
Repeat the above steps until the right pointer reaches the end of the string.

Time Complexity: O(n) where n is the length of the input string. Space Complexity: O(min(n, k)) where n is the length of the input string and k is the size of the substring.

: :

Solution Code

class Solution {
    public int numKLenSubstrNoRepeats(String S, int K) {
        int[] count = new int[26];
        int res = 0, left = 0;
        
        for (int right = 0; right < S.length(); right++) {
            count[S.charAt(right) - 'a']++;
            while (count[S.charAt(right) - 'a'] > 1) {
                count[S.charAt(left) - 'a']--;
                left++;
            }
            if (right - left + 1 == K) {
                res++;
                count[S.charAt(left) - 'a']--;
                left++;
            }
        }
        
        return res;
    }
}

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