<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 != 0:
            return False

        target = total // k
        nums.sort(reverse=True)

        n = len(nums)
        if nums[0] > target:
            return False

        @lru_cache(None)
        def backtrack(used, k, current_sum):
            if k == 0:
                return True
            if current_sum == target:
                return backtrack(used, k - 1, 0)
            for i in range(n):
                # Check if the i-th element is already used
                if (used >> i) & 1 == 0 and current_sum + nums[i] <= target:
                    # Mark the i-th element as used by setting the i-th bit
                    new_used = used | (1 << i)
                    if backtrack(new_used, k, current_sum + nums[i]):
                        return True
            return False

        # Start the backtrack function with no elements used, k subsets to fill, and initial sum of 0
        return backtrack(0, k, 0)
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