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Intuition\n\n# Approach\nArray of integers where the number representing the colour is the index. The number at each index represents the frequency. \nEach play
# Intuition\n\n# Approach\nArray of integers where the number representing the colour is the index. The number at each index represents the frequency. \nEach player "owns" one of these via key-value pairing in the HashMap.\nIterating through pick and update the array in the hashmap gives the frequency of said colour  \n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n+m)$$\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n# Code\n```java []\nclass Solution {\n    public int winningPlayerCount(int n, int[][] pick) {\n        int winners = 0;\n        Map<Integer,int[]> playerToBalls = new HashMap<>();\n        for(int i = 0; i < n; i++){\n            playerToBalls.put(i,new int[11]);\n        }\n        \n        for(int i = 0; i < pick.length; i++){\n            int currentPlayer = pick[i][0];\n            int currentColour = pick[i][1];\n            int[] arr = playerToBalls.get(currentPlayer);\n            arr[currentColour] += 1;\n        }\n\n        for(Map.Entry<Integer,int[]> entry : playerToBalls.entrySet()){\n            int min = 1 + entry.getKey();\n            for(int val : entry.getValue()){\n                if(val >= min){\n                    winners++;\n                    break;\n                }\n            }\n        }\n\n        return winners;\n    }\n}\n\n```
['Java']