LeetCode 3303: Find the Occurrence of First Almost Equal Substring Solution

Master LeetCode problem 3303 (Find the Occurrence of First Almost Equal Substring), 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.

3303. Find the Occurrence of First Almost Equal Substring

Hard

String String Matching

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will iterate through the string s using a window of size equal to the length of pattern. At each step, we will check if the current substring within the window is almost equal to the pattern by allowing at most one character to be different.

We will keep track of the number of differences between the current substring and the pattern. If the number of differences is greater than 1, we will shrink the window by incrementing the left pointer. If the number of differences is at most 1, we will update the minimum starting index of the substring that is almost equal to the pattern.

Finally, we return the smallest starting index found or -1 if no such index exists.

Time Complexity: O(n), where n is the length of the string s. Space Complexity: O(1)

Solution Code

class Solution {
    public int findTheOccurrenceOfFirstAlmostEqualSubstring(String s, String pattern) {
        int left = 0, right = 0;
        int diffCount = 0;
        int minStart = -1;
        
        while (right < s.length()) {
            if (s.charAt(right) != pattern.charAt(right)) {
                diffCount++;
            }
            while (diffCount > 1) {
                if (s.charAt(left) != pattern.charAt(left)) {
                    diffCount--;
                }
                left++;
            }
            if (right - left + 1 == pattern.length()) {
                if (minStart == -1 || left < minStart) {
                    minStart = left;
                }
            }
            right++;
        }
        
        return minStart;
    }
}

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