<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(nums)
      if total % k: return False

      target = total // k

      nums.sort(reverse=True)

      taken = [0] * len(nums)
      cache = {}

      def backtrack(i: int, count: int, currSum: int) -> bool:
        if count == k - 1: return True
        if currSum > target: return False

        token = tuple(taken)
        if token in cache: return cache[token]

        if currSum == target:
          cache[token] = backtrack(0, count + 1, 0) 
          return cache[token]
        
        for j in range(i, len(nums)):
          if not taken[j]:
            taken[j] = 1
            if backtrack(j + 1, count, currSum + nums[j]): return True
            taken[j] = 0

        cache[token] = False
        return cache[token]
    
      return backtrack(0, 0, 0)
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