LeetCode 2565: Subsequence With the Minimum Score Solution

Master LeetCode problem 2565 (Subsequence With the Minimum Score), 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.

2565. Subsequence With the Minimum Score

Hard

Two Pointers String Binary Search

Problem Explanation

Explanation

To solve this problem, we can iterate through string s and keep track of the position of characters in string t that we encounter in s. We can then calculate the score based on the positions of the characters in t that we have found in s. The minimum score would be the difference between the maximum and minimum positions of the characters in t in s.

Algorithm

Create a map to store the positions of characters in string t.
Iterate through string s and update the map with the positions of characters in t found in s.
Calculate the minimum score as the difference between the maximum and minimum positions of characters in t in s.

Time Complexity

The time complexity of this solution is O(n), where n is the length of string s or t.

Space Complexity

The space complexity of this solution is O(n), where n is the length of string t.

Solution Code

class Solution {
    public int minScore(String s, String t) {
        Map<Character, List<Integer>> positions = new HashMap<>();
        for (int i = 0; i < t.length(); i++) {
            positions.computeIfAbsent(t.charAt(i), k -> new ArrayList<>()).add(i);
        }
        
        int minScore = Integer.MAX_VALUE;
        int left = -1, right = -1;
        
        for (char c : s.toCharArray()) {
            if (positions.containsKey(c)) {
                for (int pos : positions.get(c)) {
                    if (pos > right) {
                        right = pos;
                    }
                    if (left == -1 || pos < left) {
                        left = pos;
                    }
                }
            }
            
            if (left != -1 && right != -1) {
                minScore = Math.min(minScore, right - left + 1);
            }
        }
        
        return minScore == Integer.MAX_VALUE ? 0 : minScore;
    }
}

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