<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:
        s = sum(matchsticks)
        if s % 4 != 0: return False
        edge = s // 4
        matchsticks = sorted(matchsticks)
        for ms in matchsticks:
            if ms > edge: return False
            
        se = set()
        se.add((edge,edge,edge,edge))

        def helper(a,b,c,d):
            l = sorted([a,b,c,d])
            p = (l[0],l[1],l[2],l[3])
            return p

        # print("edge=",edge)
        for i, a in enumerate(matchsticks):
            newSE = set()
            # print("a=",a, "se=",se)
            for p in se:
                # print("p=",p)
                if p[0] - a >= 0: newSE.add(helper(p[0]-a,p[1],p[2],p[3]))
                if p[1] - a >= 0: newSE.add(helper(p[0],p[1]-a,p[2],p[3]))
                if p[2] - a >= 0: newSE.add(helper(p[0],p[1],p[2]-a,p[3]))
                if p[3] - a >= 0: newSE.add(helper(p[0],p[1],p[2],p[3]-a))
            se = set()
            se = newSE
            
        if (0,0,0,0) in se: return True
        return False


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