LeetCode 3398: Smallest Substring With Identical Characters I Solution

Master LeetCode problem 3398 (Smallest Substring With Identical Characters I), a hard 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.

3398. Smallest Substring With Identical Characters I

Hard

Array Binary Search Enumeration

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We can iterate through the binary string s and maintain a window that contains only one type of character. We will keep track of the count of the other character that falls outside the window. Whenever the count exceeds the allowed number of operations numOps, we shrink the window from the left until the count is within the limit.

By doing this, we can find the minimum length of the longest substring with identical characters after performing at most numOps operations.

Initialize variables left and right for the window boundaries.
Initialize a map to count the occurrences of each character.
Iterate over the binary string s using the window approach.
Update the count of characters in the map.
If the count exceeds numOps, shrink the window from the left.
Update the maximum length of the window. :

Solution Code

class Solution {
    public int minOperations(String s, int numOps) {
        int n = s.length();
        int left = 0, right = 0, maxLen = 0, maxCount = 0;
        Map<Character, Integer> countMap = new HashMap<>();
        
        while (right < n) {
            char c = s.charAt(right);
            countMap.put(c, countMap.getOrDefault(c, 0) + 1);
            maxCount = Math.max(maxCount, countMap.get(c));
            
            while (right - left + 1 - maxCount > numOps) {
                char leftChar = s.charAt(left);
                countMap.put(leftChar, countMap.get(leftChar) - 1);
                left++;
            }
            
            maxLen = Math.max(maxLen, right - left + 1);
            right++;
        }
        
        return maxLen;
    }
}

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