<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:

        # summesh = sum(matchsticks)
        # if len(matchsticks)<4:
        #     return False
        # if summesh %4 !=0:
        #     return False
        
        # edge = summesh//4
        # @cache
        # def dfs(s1, s2, s3, s4, i):
        #     if s1==s2==s3==s4==edge:
        #         return i==len(matchsticks)

        #     if any([s>edge for s in [s1,s2,s3,s4]]):
        #         return False

        #     ret = False
        #     ret = ret or dfs(s1+matchsticks[i], s2, s3, s4, i+1)
        #     ret = ret or dfs(s1, s2+matchsticks[i], s3, s4, i+1)
        #     ret = ret or dfs(s1, s2, s3+matchsticks[i], s4, i+1)
        #     ret = ret or dfs(s1, s2, s3, s4+matchsticks[i], i+1)
        #     return ret
        
        # return dfs(0, 0, 0, 0, 0)

        n = len(matchsticks)
        totalLength = sum(matchsticks)
        sideLength = totalLength // 4
        if sideLength * 4 != totalLength: return False
        
        matchsticks.sort(reverse=True)

        @cache
        def findSquare(i, s1, s2, s3, s4):
            if i == n:
                return s1 == s2 == s3 == s4 == 0

            return ((s1 >= matchsticks[i] and findSquare(i + 1, s1 - matchsticks[i], s2, s3, s4))
                or (s2 >= matchsticks[i] and findSquare(i + 1, s1, s2 - matchsticks[i], s3, s4))
                or (s3 >= matchsticks[i] and findSquare(i + 1, s1, s2, s3 - matchsticks[i], s4))
                or (s4 >= matchsticks[i] and findSquare(i + 1, s1, s2, s3, s4 - matchsticks[i])))
        
        return findSquare(0, sideLength, sideLength, sideLength, sideLength)
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