[leetcode] 914. X of a Kind in a Deck of Cards

Description

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

分析

题目的意思是：把一个数组里面的数字平均分成k个组，每个组里面的数字是一样的，这在leetcode里面是一道easy题目，我也没做出来，太尴尬了，答案是先统计每个数的频率，然后遍历寻找合适的X，如果找到就能够分成功，找不到，就是False了。

代码

class Solution:
    def hasGroupsSizeX(self, deck: List[int]) -> bool:
        count=collections.Counter(deck)
        N=len(deck)
        for X in range(2,N+1):
            if(N%X==0):
                flag=True
                for v in count.values():
                    if(v%X!=0):
                        flag=False
                if(flag):
                    return True
        return False
        

参考文献

[LeetCode] Approach 1: Brute Force