<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
class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        target, resid = divmod(sum(nums), k)
        if resid: return False

        n = len(nums)
        nums.sort()

        @lru_cache(maxsize=None)
        def backtrack(bitmask: int) -> bool:
            if bitmask == 0:
                return 0
            
            for i in range(n):
                # the bitmask will be indexed left to right
                if not bitmask & (1 << i):
                    continue
                # print(f"{bin(bitmask)=}, {i=}")

                new_bitmask = bitmask ^ (1 << i)
                remain = backtrack(new_bitmask)
                if remain == -1:
                    continue
                if remain + nums[n-i-1] <= target:
                    # print(f"{(remain + nums[n-i-1])=}, {remain=}, {nums[n-i-1]=}")
                    return (remain + nums[n-i-1]) % target

                    
            return -1

        
        return backtrack((1 << n) - 1) == 0
60575