LeetCode - Algorithms - 2347. Best Poker Hand

Problem

2347. Best Poker Hand

Python

Counter

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from collections import Counter

class Solution:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        c = Counter(ranks).values()
        if all(suits[0] == suit for suit in suits):
            return "Flush"
        elif max(c) > 2:
            return "Three of a Kind"
        elif max(c) == 2:
            return "Pair"
        else:
            return "High Card"

Submission Detail

  • 98 / 98 test cases passed.
  • Runtime: 58 ms, faster than 42.46% of Python3 online submissions for Best Poker Hand.
  • Memory Usage: 13.8 MB, less than 97.47% of Python3 online submissions for Best Poker Hand.

Dictionary

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class Solution:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        d = dict()
        for rank in ranks:
            if rank not in d:
                d[rank] = 1
            else:
                d[rank] += 1
        c = d.values()
        if all(suits[0] == suit for suit in suits):
            return "Flush"
        elif max(c) > 2:
            return "Three of a Kind"
        elif max(c) == 2:
            return "Pair"
        else:
            return "High Card"

Submission Detail

  • 98 / 98 test cases passed.
  • Runtime: 71 ms, faster than 7.61% of Python3 online submissions for Best Poker Hand.
  • Memory Usage: 13.9 MB, less than 68.49% of Python3 online submissions for Best Poker Hand.