LeetCode 914. X of a Kind in a Deck of Cards gcd 辗转相除 reduce-优快云博客

本文探讨了一种算法，用于判断一副包含整数的牌组是否能被分成一个或多个组，每组有X张相同的牌，X大于等于2。通过实例演示了如何使用该算法进行牌组的有效分割。

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In a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

Each group has exactly X cards.
All the cards in each group have the same integer.

Example 1:

Input: deck = [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].

Example 2:

Input: deck = [1,1,1,2,2,2,3,3]
Output: false´
Explanation: No possible partition.

Example 3:

Input: deck = [1]
Output: false
Explanation: No possible partition.

Example 4:

Input: deck = [1,1]
Output: true
Explanation: Possible partition [1,1].

Example 5:

Input: deck = [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2].

Constraints:

1 <= deck.length <= 10^4
0 <= deck[i] < 10^4

-------------------------------------

from collections import Counter
from functools import reduce

class Solution:
    def hasGroupsSizeX(self, deck) -> bool:
        def gcd(x, y):
            while (y):
                x, y = y, x % y
            return x
        counter = Counter(deck)
        return reduce(gcd, counter.values()) >= 2

s = Solution()
print(s.hasGroupsSizeX(deck = [1,2,3,4,4,3,2,1]))