<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:
        total_sum = sum(nums)
        if(total_sum % k != 0):
            return False
        partition_sum = total_sum // k
        nums.sort()
        memory = {}

        def aux(start, subsets_found, curr_sum, mask):
            if(subsets_found == k):
                return True
            if(mask in memory):
                return memory[mask]
            if(curr_sum == partition_sum):
                return aux(0, subsets_found + 1, 0, mask)
            if(start >= len(nums)):
                return False
            
            for i in range(start, len(nums)):
                if(nums[i] + curr_sum > partition_sum):
                    break
                if( mask & (1 << i) == 0 ):
                    if(aux(i + 1, subsets_found, curr_sum + nums[i], mask | (1 << i))):
                        return True
            memory[mask] = False
            return False
        
        return aux(0, 0, 0, 0)
        
34010