<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:
        n = len(nums)
        total = sum(nums)
        if total%k!=0:
            return False
        target = total//k
        nums.sort()
        taken = ['0']*n
        memo = {}
        
        def backtrack(index, count, sum_):
            taken_str = ''.join(taken)
            if count == k:
                return True
            if sum_ > target:
                return False
            if taken_str in memo:
                return memo[taken_str]
            if sum_ == target:
                memo[taken_str] = backtrack(0, count+1, 0)
                return memo[taken_str]
            for j in range(index, n):
                if taken[j]=='0':
                    taken[j] = '1'
                    if backtrack(j+1, count, sum_+nums[j]):
                        return True
                    taken[j] = '0'
            memo[taken_str] = False        
            return memo[taken_str]        
        return backtrack(0,0,0)            
43670