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

<p>Return <code>true</code> if you can make this square and <code>false</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/09/matchsticks1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> matchsticks = [1,1,2,2,2]
<strong>Output:</strong> true
<strong>Explanation:</strong> You can form a square with length 2, one side of the square came two sticks with length 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matchsticks = [3,3,3,3,4]
<strong>Output:</strong> false
<strong>Explanation:</strong> You cannot find a way to form a square with all the matchsticks.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= matchsticks.length &lt;= 15</code></li>
	<li><code>1 &lt;= matchsticks[i] &lt;= 10<sup>8</sup></code></li>
</ul>

class Solution:
    def makesquare(self, matchsticks: List[int]) -> bool:
        nums=matchsticks
        if sum(matchsticks)%4!=0:
            return False
        t=[sum(matchsticks)//4]
        dp={}
        def dfs(mask,subsum,k):
            if mask in dp:
                return dp[mask]
            if subsum==0:
                subsum=t[0]
                k-=1
            if k==0:
                dp[mask]=True
                return True

            for i in range(len(nums)):
                if 1<<i & mask:
                    continue
                if nums[i]>subsum:
                    dp[mask|1<<i]=False
                else:
                    if dfs(mask|1<<i,subsum-nums[i],k):
                        dp[mask]=True
                        return dp[mask]
            dp[mask]=False
            return False
        return dfs(0,t[0],4)
        
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