<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> if it is possible to divide this array into <code>k</code> non-empty subsets whose sums are all equal.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,2,3,5,2,1], k = 4
<strong>Output:</strong> true
<strong>Explanation:</strong> It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 3
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 16</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li>The frequency of each element is in the range <code>[1, 4]</code>.</li>
</ul>

class Solution:

    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        n = len(nums)
        nums.sort()
        s = sum(nums)
        v = s // k
        
        if s % k > 0 or nums[-1] > v:
            return False
        
        visited = set()
        
        def bt(x, curr, cnt): 
            if x in visited:
                return False
            
            visited.add(x)

            if curr == 0:
                if cnt == 0:
                    return True
                else:
                    cnt -= 1
                    curr = v

            
            for i in range(n):
                if (x >> i) & 1 == 0:
                    continue

                if nums[i] > curr:
                    return False            

                if bt(x ^ (1 << i), curr - nums[i], cnt):
                    return True
            
            return False
        
        return bt((1 << n) - 1, v, k - 1)
                    

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