<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 target != int(target):
            return False
        
        used = ['0'] * len(nums)
        checked = set()
        nums.sort()
        def dfs(s, count, index):
            if s == target:
                if count == len(nums):
                    return True
                return dfs(0, count, 0)
            status = ''.join(used)
            if status in checked:
                return False
            for i in range(index, len(nums)):
                if used[i] == '1':
                    continue
                if s+nums[i] > target:
                    checked.add(status)
                    return False
                used[i] = '1'
                if dfs(s+nums[i], count+1, i+1):
                    return True
                used[i] = '0'
            checked.add(status)
            return False
        
        return dfs(0, 0, 0)
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