Find the Number of Winning Players

You are given an integer n representing the number of players in a game and a 2D array pick where pick[i] = [x<sub>i</sub>, y<sub>i</sub>] represents that the player x<sub>i</sub> picked a ball of color y<sub>i</sub>.

Player i wins the game if they pick strictly more than i balls of the same color. In other words,

• Player 0 wins if they pick any ball.

• Player 1 wins if they pick at least two balls of the same color.

• ...

• Player i wins if they pick at leasti + 1 balls of the same color.

Return the number of players who win the game.

Note that multiple players can win the game.

LeetCode Problem - 3238

class Solution {
    public int winningPlayerCount(int n, int[][] pick) {
        // Initialize the answer to keep track of the number of winning players
        int answer = 0;

        // Create a 2D array to count the number of times each player picks each color
        int[][] colorCounts = new int[n][11]; // Assuming there are 10 colors (index 1 to 10)

        // Iterate through each pick and increment the corresponding player's color count
        for(int[] p : pick) {
            int player = p[0]; // Player index
            int color = p[1];  // Color index
            colorCounts[player][color]++;
        }

        // Iterate through each player's color count to determine if they are a winning player
        for(int i = 0; i < n; i++) {
            for(int val : colorCounts[i]) {
                // A player is considered a winner if they have picked a color more times than their index
                if(val > i) {
                    answer++; // Increment the count of winning players
                    break;    // Stop checking further colors for this player
                }
            }
        }

        // Return the total number of winning players
        return answer;
    }
}