Python 3 || 2 lines, Counter  ||
\n\nclass Solution:\n    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:\n\n        ctr = Counter(map(tuple,pick))\n\n        return len({pl
\n```\nclass Solution:\n    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:\n\n        ctr = Counter(map(tuple,pick))\n\n        return len({player for (player,_), count in ctr.items() if count > player})\n```\n[https://leetcode.com/problems/find-the-number-of-winning-players/submissions/1343258032/](https://leetcode.com/problems/find-the-number-of-winning-players/submissions/1343258032/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(*N*), in which *N* ~ `len(pick)`.
['Python3']