Description

You are given an integer array matchsticks where matchsticks[i] is the length of the ith matchstick. You want to use all the matchsticks to make one square. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Return true if you can make this square and false otherwise.

Example

Input: matchsticks = [1,1,2,2,2]
Output: true
Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

Code

class Solution(object):
    def makesquare(self, arr):
        """
        :type matchsticks: List[int]
        :rtype: bool
        """
        rec_len = sum(arr) // 4
        if sum(arr) % 4 > 0:
            return False 
        h = {0:0}
        candidate = []
        seen = set()
       
        if any([v > rec_len for v in arr]):
            return False
        
        bits = sum([(1 << i) for i in range(len(arr))])
        
        for i in range(len(arr)):    # 先把满足边的组合计算出来，用于后续迭代。耗时 369 ms， 战胜了 93 % 的提交。不用这个方法会 TLE 的
                                           # 采用 dfs 的话，代码很容易实现。但是耗时会变成 1500 ms + 
            tmp = {}
            for k in h:
                if rec_len > h[k] + arr[i]:
                    tmp[k | (1 << i)] = h[k] + arr[i]
                elif rec_len == h[k] + arr[i]:
                    candidate.append(k | (1 << i))
            h.update(tmp) 
            
        h = {0:0}
        for v in candidate:
            tmp = {}
            for s in h:
                if s & v == 0:
                    if s | v == bits:
                        return True
                    tmp[s | v] = 0
            h.update(tmp)
        return False