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LeetCode 3455: Shortest Matching Substring Solution

Master LeetCode problem 3455 (Shortest Matching 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.

3455. Shortest Matching Substring

Hard
Two PointersStringBinary SearchString Matching

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We iterate over the string s and for each character we encounter, we check if it matches the pattern p. We keep track of the positions of the two * characters in the pattern p and use them to determine the window of characters in s that can match the pattern.

Algorithm:

  1. Initialize two pointers p1 and p2 to store the positions of the two * characters in pattern p.
  2. Initialize variables matchStart and matchEnd to store the start and end indices of the current matching substring in s.
  3. Iterate over the string s:
    • If the character at the current index matches the character at position p1 or p2 in pattern p, update matchEnd to the current index.
    • If the characters don't match, update matchStart to the next index after the last * character.
    • Check if the current substring from matchStart to matchEnd matches the pattern p by considering the characters between the two * characters.
    • If it matches, update the length of the shortest matching substring if needed.
  4. Return the length of the shortest matching substring or -1 if no matching substring is found.

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

:

Solution Code

class Solution {
    public int shortestMatchingSubstring(String s, String p) {
        int p1 = -1, p2 = -1;
        for (int i = 0; i < p.length(); i++) {
            if (p.charAt(i) == '*') {
                if (p1 == -1) {
                    p1 = i;
                } else {
                    p2 = i;
                }
            }
        }
        
        int matchStart = 0, matchEnd = 0;
        int shortestLength = Integer.MAX_VALUE;
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == p.charAt(p1) || s.charAt(i) == p.charAt(p2)) {
                matchEnd = i;
            } else {
                matchStart = Math.max(matchStart, matchEnd + 1);
            }
            
            if (matchStart <= matchEnd && matchEnd < s.length() - 1) {
                String substring = s.substring(matchStart, matchEnd + 1);
                if (substring.contains(p.substring(p1 + 1, p2))) {
                    shortestLength = Math.min(shortestLength, matchEnd - matchStart + 1);
                }
            }
        }
        
        return shortestLength == Integer.MAX_VALUE ? -1 : shortestLength;
    }
}

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