<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:
        if sum(nums) % k != 0:
            return False
        
        target = sum(nums) // k
        dp = {}
        def dfs(visited, cursum, count):
            if count == k-1:
                return True
            
            if tuple(visited) in dp:
                return dp[tuple(visited)]
            
            for i in range(len(visited)):
                if cursum + visited[i] == target:
                    tmp = visited[:i] + visited[i+1:]
                    if dfs(tmp, 0, count + 1):
                        dp[tuple(visited)] = True
                        return True
                    
                if cursum + visited[i] < target:
                    tmp = visited[:i] + visited[i+1:]
                    if dfs(tmp, cursum + visited[i], count):
                        dp[tuple(visited)] = True
                        return True
                    
            dp[tuple(visited)] = False
            return False
        
        return dfs(nums, 0, 0)
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