<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 recur(self, nums, k, target, i, visited):
        state = (k,target,visited)
        if state in self.memo:
            return self.memo[state]
        if target == 0:
            k -= 1
            if k == 0:
                return True
            target = self.fresh
            return self.recur(nums, k, target, 0, visited)
        
        for j in range(i, len(nums)):
            if visited & (1 << j):
                 continue
            if nums[j] > target:
                continue
            
            visited |= (1 << j)
            if self.recur(nums, k, target - nums[j], j + 1, visited):
                self.memo[state] = True
                return True
            visited ^= (1 << j)
        self.memo[state] = False
        return False
    
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        sumi = sum(nums)
        visited = [-1] * len(nums)
        
        if sumi % k != 0:
            return False
        
        target = sumi // k
        self.fresh = target
        
        nums.sort(reverse=True)
        if(nums[0]>target):
            return False
        self.memo = {}
        return self.recur(nums, k, target, 0, 0)

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