Easy to understand Solution | O(1) extra space | Beats 100% | JAVA
\n# Approach\nCount Item Picks for Each Player:\nFor each player i, create an array col[] of size 11 to count how many times the player picked each item. Loop t
\n# Approach\nCount Item Picks for Each Player:\nFor each player i, create an array col[] of size 11 to count how many times the player picked each item. Loop through the pick array, and whenever the player i picks an item j, increment col[j].\n\nCheck the Condition:\nAfter counting the picks for the player, check if any item j is picked more than i times (i.e., col[j] > i). If true for any item, increase ans by 1 because this player qualifies as a "winning" player.\n\nReset the Count:\nAfter processing each player, reset the col[] array to ensure no leftover counts affect the next player.\n\nReturn the Result:\nFinally, return ans, which is the total count of players who met the condition.\n# Complexity\n- Time complexity: $$O(n * m)$$ \nn is the number of players and m is the number of picks.\n\n- Space complexity: $$O(1)$$ extra space\n\n# Code\n```java []\nclass Solution {\n    public int winningPlayerCount(int n, int[][] pick) {\n        int ans = 0;\n        for(int i =0;i<n;i++){\n            int col[]= new int[11];\n            for(int j =0;j<pick.length;j++){\n                if(pick[j][0]==i){\n                    col[pick[j][1]]++;\n                }\n            }\n            for(int k =0;k<col.length;k++){\n                if(col[k]>i){\n                    ans++;\n                    break;\n                }\n            }\n            Arrays.fill(col, 0);\n\n        }\n        return ans;\n    }\n}\n```
['Array', 'Counting', 'Java']