<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:
        sum_nums = sum(nums)
        if sum_nums % k != 0:
            return False

        target = sum_nums // k
        memo = {}
        used = [False] * len(nums)
        nums.sort(reverse=True)                 
        
        def backtrack(i, k, curr_sum):
            key = tuple(used)
            if key in memo:
                return memo[key]

            if k == 1:
                return True

            if curr_sum == target:
                res = backtrack(0, k - 1, 0)
                memo[tuple(used)] = res
                return res

            for j in range(i, len(nums)):
                if not used[j] and curr_sum + nums[j] <= target:
                    used[j] = True
                    if backtrack(j + 1, k, curr_sum + nums[j]):
                        return True
                    used[j] = False
            
            memo[tuple(used)] = False
            return False
        
        return backtrack(0, k, 0)


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