<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 cache


class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        if len(nums) < k or sum(nums) / k != int(sum(nums) / k):
            return False

        @cache
        def dp(mask, cur):
            if mask == 0:
                return cur == 0

            if cur == 0:
                return dp(mask, sum(nums) / k)

            for bit in range(len(nums)):
                if mask & (1 << bit):
                    if nums[bit] <= cur and dp(mask ^ (1 << bit), cur - nums[bit]):
                        return True

            return False

        return dp(2 ** len(nums) - 1, sum(nums) / k)

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