LeetCode 1626: Best Team With No Conflicts Solution

Master LeetCode problem 1626 (Best Team With No Conflicts), 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.

1626. Best Team With No Conflicts

Medium

Array Dynamic Programming Sorting

Problem Explanation

Explanation:

To solve this problem, we can sort the players based on their ages and then apply dynamic programming. We sort the players based on ages so that we can avoid conflicts between younger and older players. We then iterate over each player and calculate the maximum score that can be achieved by including that player in the team. To calculate this, we compare the score of the current player with all previous players who are younger than the current player and update the maximum score accordingly. Finally, we return the maximum overall score that can be achieved.

Algorithm:

Sort the players based on their ages.
Initialize an array dp of size equal to the number of players, where dp[i] represents the maximum overall score that can be achieved by including player i.
Iterate over each player and update dp[i] by comparing the score of the current player with all previous players who are younger and update the maximum score accordingly.
Return the maximum value in the dp array as the result.

Time Complexity: O(n^2) where n is the number of players. Space Complexity: O(n) for the dp array.

Solution Code

class Solution {
    public int bestTeamScore(int[] scores, int[] ages) {
        int n = scores.length;
        int[][] players = new int[n][2];
        for (int i = 0; i < n; i++) {
            players[i] = new int[]{ages[i], scores[i]};
        }
        Arrays.sort(players, (a, b) -> (a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]));
        
        int[] dp = new int[n];
        for (int i = 0; i < n; i++) {
            dp[i] = players[i][1];
            for (int j = 0; j < i; j++) {
                if (players[i][1] >= players[j][1]) {
                    dp[i] = Math.max(dp[i], dp[j] + players[i][1]);
                }
            }
        }
        
        return Arrays.stream(dp).max().getAsInt();
    }
}

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