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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards
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本人的代码实在丑陋，这道题 LeetCode 官方答案深得我心，特意记录下来。
class Solution:
    def hasGroupsSizeX(self, deck: List[int]) -> bool:
        # version 1:
        # from fractions import gcd
        # vals = collections.Counter(deck).values()
        # return reduce(gcd,vals) >= 2
        # version 2:
        d_counter = collections.Counter(deck)
        N = len(deck)
        for X in range(2, N + 1):
            if N % X == 0:
                if all(v % X == 0 for v in d_counter.values()):
                    return True
        return False