<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:
        n = len(matchsticks)
        total = sum(matchsticks)
        target = total/4
        if n < 4 or total % 4 != 0:
            return False
        matchsticks = sorted(matchsticks, reverse=True)

        @cache
        def dfs(sides, idx):
            if idx == n:
                if all([sides[i] == target for i in range(4)]):
                    return True
                return False
            curr = matchsticks[idx]
            sides = list(sides)
            for i in range(4):
                if sides[i] + curr <= target:
                    sides[i] += curr
                    if dfs(tuple(sides), idx+1):
                        return True
                    sides[i] -= curr
            return False

        return dfs((0,0,0,0), 0)
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