<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 dfs(self, rounds: int, cur_idx: int, cur_sum: int, mask: int, k: int) -> bool:
        if cur_idx == len(self.nums):
            if cur_sum != self.target:
                return False
            elif rounds + 1 == k:
                return True
            else:
                res = self.dfs(rounds+1, 0, 0, mask, k)
                self.cache[mask] = res
                return res
        elif mask in self.cache:
            return self.cache[mask]
        elif cur_sum > self.target:
            return False
        else:
            res = False
            tmp1 = self.dfs(rounds, cur_idx+1, cur_sum, mask, k)
            res = res or tmp1
            if (mask & (1<<cur_idx)) == 0:
                next_mask = (mask | (1<<cur_idx))
                tmp2 = self.dfs(rounds, cur_idx+1, cur_sum+self.nums[cur_idx], next_mask, k)
                res = res or tmp2
            self.cache[(rounds, mask)] = res
            return res
    
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        if sum(nums) % k != 0:
            return False
        self.target = sum(nums) // k
        self.nums = sorted(nums, reverse=True)
        self.terminal = 2**len(nums)-1
        self.cache = {}
        
        res = self.dfs(0, 0, 0, 0, k)
        
        return res
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