Counting | O(𝑛𝑦+𝑚), O(𝑛×𝑦) | 10.00%, 31.25%
Intuition\n Describe your first thoughts on how to solve this problem. \nWe note that the question is generous with its constraints, in that there are up to 10 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe note that the question is generous with its constraints, in that there are up to 10 players (`2 <= n <= 10`) and up to 11 colours of balls (`0 <= y <= 10`). As such, our approach performs <ins>counting</ins> as it iterates through the picks  sequentially, adds the pick to the respective ball count, and checks against the winning conditions for each player.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Initialize a list of dictionaries to track counts of colors picked by each player.\n- Use a set to track which players have already won.\n- Iterate through each pick and update the color counts for the respective player.\n- Check if the player meets their winning condition and update the count of winning players accordingly.\n# Complexity\n*where $m$ is length of the array `pick`, $n$ is number of players, and $y$ is number of colours.*\n- **Time complexity: O(\uD835\uDC5By+\uD835\uDC5A)**\nTime is taken to create the list of dictionaries, and for processing of the picks, where each pick is processed exactly once.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- **Space complexity: O(\uD835\uDC5B\xD7\uD835\uDC66)**\nThe list of dictionaries requires space for each player and the respective counts for each color.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n# Code\n| Runtime    | Memory |\n| -------- | ------- |\n| **164** ms &#124; beats **10.00%** | **16.64** MB &#124; beats **31.25%**|\n```python3 []\nclass Solution:\n    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:\n        mylist = []\n        # mydict = {0:0,1:0,2:0,3:0,4:0,5:0,6:0,7:0,8:0,9:0,10:0}\n        player_counted = set()\n        count = 0\n        for i in range(n):\n            mylist.append({0:0,1:0,2:0,3:0,4:0,5:0,6:0,7:0,8:0,9:0,10:0})\n        for x in pick:\n            # print(mylist)\n            if x[0] in player_counted:\n                continue\n            if mylist[x[0]][x[1]] >= x[0]:\n                count += 1\n                player_counted.add(x[0])\n            else:\n                mylist[x[0]][x[1]] += 1\n        return count\n```\n# Detailed Approach\n1. Initialization:\n    - Create a list called `mylist` where each element is a dictionary that maps colors to their counts for each player.\n    - Create a set called `player_counted` to keep track of players who have already won.\n    - Initialize a counter `count` of number of players won to zero.\n2. Processing Picks:\n    - For each pick, check if the player has already been counted as a winner (in`player_counted`set). If so, skip to the next pick.\n    - If not, check if the count of the color picked meets or exceeds the winning threshold (`mylist[x[0]][x[1]] >= x[0]`).\n    - If the condition is satisfied, increment the `count` of winning players and add the player to the `player_counted` set.\n    - Otherwise, increment the count of that color in the player\'s dictionary.\n3. Return the Count:\n    - At the end of processing all picks, return the total `count` of winning players.\n\n# Detailed Complexity Breakdown\n### Time Complexity: $O(ny+m)$\n1. Initializing mylist: \n    - We create a dictionary for each of the $$n$$ players, where each dictionary contains $y$ colours, taking $O(ny)$ time.\n2. Processing each pick: \n    - Processing $m$ number of picks takes $O(m)$ time.\n        - Each color count update and check involves dictionary operations which are $O(1)$ for insertion and retrieval.\n\n### Space Complexity: $O(n\\times y)$\n1. Space for `mylist`: \n    - As we create a dictionary containing all $c$ colours for each of $n$ players, space complexity is $O(n\\times y)$.\n2. Space for `player_counted`: \n    - In the worst case, if all $n$ players are counted as winners, the `player_counted` set will take $O(n)$ space.\n***\nCheck out my solutions to other problems here! \uD83D\uDE03\n*https://leetcode.com/discuss/general-discussion/5942132/my-collection-of-solutions*\n***
['Counting', 'Python3']