<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:
        s = sum(nums)
        avg = s //k
        if s % k != 0:
            return False  
        
        
        visited = set()
        cache = {}
        nums.sort(reverse=True)

        def backtrack(start, k , total, visited):
            
            key = tuple(sorted(visited))
            if (start, k, total, key ) in cache:
                return cache[(start, k, total, key )]

            if k == 0:
                return True
            
            if total == avg:
                return backtrack(0, k-1, 0, visited)
            
            if start == len(nums):
                return
            

            for i in range(start, len(nums)):
                if i in visited:
                    continue 
                
                if total + nums[i] > avg:
                    continue 
                
                visited.add(i)
                if backtrack(i+1, k, total + nums[i], visited):
                    key = tuple(sorted(visited))
                    cache[(start, k, total, key )] = True
                    return True
                visited.remove(i)
            key = tuple(sorted(visited))
            cache[(start, k, total, key )] = False
            return False 
        
        return backtrack(0, k, 0, visited)

        
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