<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:
        target = sum(nums)//k 
        if sum(nums)%k  != 0:
            return False
        
        self.memo ={}
        self.used= [False]*len(nums)
        return self.backtrack(nums, target, k , 0, 0)
    
    def backtrack(self, nums, target, k , start, partition):
        memo_key = tuple(self.used)

        if memo_key in self.memo:
            return self.memo[memo_key]
        if k == 0:
            self.memo[memo_key] =True
            return True 
        if target == partition:
            return self.backtrack(nums, target, k-1, 0, 0)
        
        for i in range(start, len(nums)):
            if self.used[i]:continue
            if partition + nums[i] > target: continue 

            self.used[i] = True 
            if self.backtrack(nums, target, k, i+1, partition+ nums[i]):
                self.memo[memo_key] =True 
                return True

            self.used[i] =False
        self.memo[memo_key] = False
        return False
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