<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)
        s = sum(nums)
        if s % k != 0:
            return False
        tgt = s // k
        nums.sort ()
        
        @lru_cache(None)
        def check (bits, cursum, curk):
            if bits == ((1<<N)-1) and curk == 0:
                return True
            
            for j in range (N):
                if bits & (1<<j):
                    continue
                    
                if cursum + nums[j] > tgt:
                    break
                    
                bits |= (1<<j)
                if cursum + nums[j] == tgt:
                    ret = check (bits, 0, curk-1)
                else:
                    ret = check (bits, cursum+nums[j], curk)
                if ret:
                    return True
                bits &= ~(1<<j)                
                
            return False
        
        return check (0, 0, k)
                        
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