<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:
        if len(nums) < k or int(sum(nums) / k) != sum(nums) / k:
            return False
        nums.sort(reverse=True)
        def dp(mask, curr, memo):
            if mask == 0:
                return curr == 0
            elif curr == 0:
                return dp(mask, sum(nums) / k, memo)
            if (mask, curr) not in memo:
                res = False
                for bit in range(len(nums)):
                    if mask & (1 << bit):
                        if nums[bit] > curr:
                            continue
                        if dp(mask ^ (1 << bit), curr - nums[bit], memo):
                            res = True
                            break
                memo[(mask, curr)] = res
            return memo[(mask, curr)]
        return dp(2 ** len(nums) - 1, sum(nums) / k, dict())
23143