Detailed Explanation of approach & Code - 100% beats
Summary\n- The problem requires determining how many players win based on their ball-picking pattern. The key insight is that a player wins if they have picked 
# Summary\n- The problem requires determining how many players win based on their ball-picking pattern. The key insight is that a player wins if they have picked more balls of any single color than their index. By counting the frequency of each color picked by each player, we can easily check if a player meets the winning condition.\n\n# Approach\n1. Initialization: Create a 2D array where the rows represent players and the columns represent the count of each color (0-10) picked by the player.\n2. Counting: Iterate through the given picks and update the 2D array to reflect the number of each color picked by each player.\n3. Winning Check: For each player, check if any color count exceeds their index. If it does, the player wins.\n4. Result: Count the number of players who meet the winning condition.\n\n# Complexity\n- Time complexity: O(m+n\xD7c), where m is the length of the pick array, n is the number of players, and c is the number of colors (which is a constant 11).\n\n- Space complexity:  O(n\xD7c), where n is the number of players and c is the number of colors (11), for storing the color counts for each player.\n\n# Code\n\n```Java []\nclass Solution {\n    public int winningPlayerCount(int n, int[][] pick) {\n        // Step 1: Create a 2D array to count colors picked by each player\n        int[][] colorCounts = new int[n][11];\n\n        // Step 2: Update the color counts based on the picks\n        for (int[] p : pick) {\n            int player = p[0], color = p[1];\n            colorCounts[player][color]++;\n        }\n\n        // Step 3: Count the number of winning players\n        int count = 0;\n        for (int i = 0; i < n; i++) {\n            for (int val : colorCounts[i]) {\n                if (val > i) {\n                    count++;\n                    break;\n                }\n            }\n        }\n\n        return count;\n    }\n}\n```\n```C++ []\nclass Solution {\npublic:\n    int winningPlayerCount(int n, vector<vector<int>>& pick) {\n        // Step 1: Create a 2D array to count colors picked by each player\n        vector<vector<int>> colorCounts(n, vector<int>(11, 0));\n\n        // Step 2: Update the color counts based on the picks\n        for (const auto& p : pick) {\n            int player = p[0], color = p[1];\n            colorCounts[player][color]++;\n        }\n\n        // Step 3: Count the number of winning players\n        int count = 0;\n        for (int i = 0; i < n; ++i) {\n            for (int val : colorCounts[i]) {\n                if (val > i) {\n                    count++;\n                    break;\n                }\n            }\n        }\n\n        return count;\n    }\n};\n```\n```Python []\nfrom typing import List\n\nclass Solution:\n    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:\n        # Step 1: Create a 2D list to count colors picked by each player\n        color_counts = [[0] * 11 for _ in range(n)]\n\n        # Step 2: Update the color counts based on the picks\n        for player, color in pick:\n            color_counts[player][color] += 1\n\n        # Step 3: Count the number of winning players\n        count = 0\n        for i in range(n):\n            if any(val > i for val in color_counts[i]):\n                count += 1\n\n        return count\n```\n```Javascript []\n/**\n * @param {number} n\n * @param {number[][]} pick\n * @return {number}\n */\nvar winningPlayerCount = function(n, pick) {\n    // Step 1: Create a 2D array to count colors picked by each player\n    const colorCounts = Array.from({ length: n }, () => Array(11).fill(0));\n\n    // Step 2: Update the color counts based on the picks\n    for (const [player, color] of pick) {\n        colorCounts[player][color]++;\n    }\n\n    // Step 3: Count the number of winning players\n    let count = 0;\n    for (let i = 0; i < n; i++) {\n        if (colorCounts[i].some(val => val > i)) {\n            count++;\n        }\n    }\n\n    return count;\n};\n```\n```Dart []\nclass Solution {\n  int winningPlayerCount(int n, List<List<int>> pick) {\n    // Step 1: Create a 2D list to count colors picked by each player\n    List<List<int>> colorCounts = List.generate(n, (_) => List.filled(11, 0));\n\n    // Step 2: Update the color counts based on the picks\n    for (List<int> p in pick) {\n      int player = p[0], color = p[1];\n      colorCounts[player][color]++;\n    }\n\n    // Step 3: Count the number of winning players\n    int count = 0;\n    for (int i = 0; i < n; i++) {\n      if (colorCounts[i].any((val) => val > i)) {\n        count++;\n      }\n    }\n\n    return count;\n  }\n}\n```\n```Golang []\nfunc winningPlayerCount(n int, pick [][]int) int {\n    // Step 1: Create a 2D slice to count colors picked by each player\n    colorCounts := make([][]int, n)\n    for i := range colorCounts {\n        colorCounts[i] = make([]int, 11)\n    }\n\n    // Step 2: Update the color counts based on the picks\n    for _, play := range pick {\n        player, color := play[0], play[1]\n        colorCounts[player][color]++\n    }\n\n    // Step 3: Count the number of winning players\n    count := 0\n    for i := 0; i < n; i++ {\n        for _, val := range colorCounts[i] {\n            if val > i {\n                count++\n                break\n            }\n        }\n    }\n\n    return count\n}\n```
['Array', 'Counting', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart']