<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:
        def find(mask,k):
            total = 0
            for bit in range(n):
                if mask&(1<<bit) == 0: total += nums[bit]
            if total and total%avg == 0: k -= 1
            if k == 0: return True
            if dp[mask][k] != -1: return dp[mask][k]
            remaining = avg - total%avg
            res = False
            for bit in range(n):
                if mask&(1<<bit) and nums[bit] <= remaining:
                    if find(mask^(1<<bit),k):
                        res = True
                        break
            dp[mask][k] = res
            return res
        n = len(nums)
        s = sum(nums)
        avg = s//k
        if s%k != 0 or max(nums) > avg: return False
        dp = [[-1]*k for _ in range(1<<n)]
        return find((1<<n)-1,k-1)
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