<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:
        total = sum(matchsticks)

        if total % 4 != 0:
            return False

        s = total // 4

        if max(matchsticks) > s:
            return False
        
        # matchsticks.sort(reverse=True)

        # sides = [0] * 4

        # def help(i):
        #     if i == len(matchsticks) and all(side == s for side in sides):
        #         print(i, sides)
        #         return True
        #     if any(side > s for side in sides):
        #         return False
            
        #     for i in range(len(matchsticks)):
        #         for j in range(len(sides)):
        #             sides[j] += matchsticks[i]
        #             if help(i + 1):
        #                 return True
        #             sides[j] -= matchsticks[i]
        #         if not sides[j]:
        #             break
            
        #     return False

        # return help(0)

        matchsticks.sort(reverse=True)

        @lru_cache(None)
        def dp(used, side):
            if side == 0:
                side = s
            if used == (1 << len(matchsticks)) - 1:
                return True

            for i in range(len(matchsticks)):
                if used & (1 << i) or matchsticks[i] > side:
                    continue
                if dp(used | (1 << i), side - matchsticks[i]):
                    return True
            
            return False

        return dp(0, s)
26072