easy||using dic,list,set methods
\n\n\n# Approach\n1. add the elements after converting it into a string to count the number of occurances\n2. then by taking the x[i] as in the player from stri
![image.png](https://assets.leetcode.com/users/images/4dfea6b9-21c1-4961-975d-284dfaf508fc_1727515369.377939.png)\n\n\n# Approach\n1. add the elements after converting it into a string to count the number of occurances\n2. then by taking the x[i] as in the player from string and finding if its had more than it\'s destines player no\n3. if so they are added to a set to remove dupliates\n\n\n# Code\n```python3 []\nclass Solution:\n    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:\n        d={}\n        for x in pick:\n            if str(x) not in d:\n                val=1\n            else:\n                val=d[str(x)]+1\n            d.update({str(x):val})\n        print(d)\n        st=set({})\n        for x in d.keys():\n            if int(x[1])<d[x]:\n                print(x,d[x])\n                st.add(int(x[1]))\n        return len(st)\n        \n```
['Python3']