<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>

from functools import lru_cache
from typing import List

class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        
        @lru_cache(None)
        def backtrack(num_subsets, subset_sum, used):
            if not num_subsets:
                return True
            
            if subset_sum == target_sum:
                return backtrack(num_subsets - 1, 0, used)
            
            for i in range(len(nums)):
                if not used[i] and subset_sum + nums[i] <= target_sum:
                    # Create a new tuple with the updated state
                    new_used = used[:i] + (True,) + used[i+1:]
                    if backtrack(num_subsets, subset_sum + nums[i], new_used):
                        return True

            return False

        if sum(nums) % k != 0:
            return False
        
        nums.sort(reverse=True)
        target_sum = sum(nums) / k
        
        # Early termination check
        if nums[0] > target_sum:
            return False
        
        return backtrack(k, 0, tuple([False] * len(nums)))

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