Beats 90%
Intuition\nThe problem asks us to determine how many players have won based on a list of picks. A player wins if the number of times a certain target is picked 
# Intuition\nThe problem asks us to determine how many players have won based on a list of picks. A player wins if the number of times a certain target is picked by them exceeds their player number, and this should happen only once for each player.\n\n# Approach\n1. Create a 2D vector `v` of size 11x11 to store the count of times each player picks each target.\n2. Create a 1D vector `won` of size 11 to keep track of whether a player has already won or not.\n3. Iterate through each pick in the `pick` vector:\n   - Increment the count of the corresponding player and target in the 2D vector `v`.\n   - If the count exceeds the player\'s number and the player hasn\'t already won, increase the win count (`ans`) and mark the player as won.\n4. Return the total count of winning players.\n\n# Complexity\n- **Time complexity**: $$O(m)$$ where \\(m\\) is the number of picks. Each pick is processed in constant time.\n- **Space complexity**: $$O(1)$$ as the space used for the vectors `v` and `won` is constant (fixed size of 11x11 and 11 respectively).\n\n# Code\n```cpp\nclass Solution {\npublic:\n    int winningPlayerCount(int n, vector<vector<int>>& pick) {\n        vector<vector<int>> v(11, vector<int>(11, 0));\n        int ans = 0;\n        vector<int> won(11, 0);\n\n        for (auto& i : pick) {\n            v[i[0]][i[1]]++;\n            if (v[i[0]][i[1]] > i[0] && won[i[0]] == 0) {\n                ans++;\n                won[i[0]] = 1;\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']