<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
        n = len(matchsticks)
        @cache
        def dfs(s1, s2, s3, s4, i):
            if i==n:
                return s1==s2==s3==s4==0
            
            return ((s1 >= matchsticks[i] and dfs(s1-matchsticks[i], s2, s3, s4, i+1))
                or (s2 >= matchsticks[i] and dfs(s1, s2-matchsticks[i], s3, s4, i+1))
                or (s3 >= matchsticks[i] and dfs(s1, s2, s3-matchsticks[i], s4, i+1))
                or (s4 >= matchsticks[i] and dfs(s1, s2, s3, s4-matchsticks[i], i+1)))
        
        matchsticks.sort(reverse=True)
        return dfs(edge,edge,edge,edge, 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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