<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:
        nums = matchsticks

        total = sum(nums)
        if total % 4: return False
        length = total // 4

        dp = [-1] * (1 << (len(nums) + 1))
        def dfs(used, edge, len):
            # nonlocal ans
            if edge == 0:
                return used == 0

            if (dp[used] != -1): 
                return dp[used]
            
            ans = False
            
            for i, v in enumerate(nums):
                if (used & (1 << i)) == 0: continue
                if len - v < 0: continue
                if len - v == 0:
                    ans = dfs((used ^ (1 << i)), edge - 1, length)
                else:
                    ans = dfs((used ^ (1 << i)), edge, len - v)
                
                if ans == True: break
            dp[used] = ans
            return ans
        
        used = (1 << (len(nums))) - 1
        
        return dfs(used, 4, length)
45817