<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 #[1]
        N = len(nums)
        nums.sort(reverse=True) #[2]
        
        def dp(mask, cur, memo): # Top-down DP with memoization
            if mask == 0: return cur == 0 #[3]
            elif cur == 0: return dp(mask, sum(nums)/k, memo) #[4]
            if (mask, cur) not in memo:
                res = False
                for bit in range(N): #[5]
                    if mask & (1 << bit): #[6]
                        if nums[bit] > cur: continue # Writing this to be more explicit, for easy-understanding
                        if dp(mask ^ (1 << bit), cur-nums[bit], memo): #[7]
                            res = True
                            break
                memo[(mask, cur)] = res
            return memo[(mask, cur)] 
        return dp(2**N-1, sum(nums)/k, dict()) #[8]
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