<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)
        if n == 1:
            return k == 1
        
        total = sum(nums)
        if total % k != 0:
            return False
        
        target = total // k
        # visited = [False] * n
        # memorize state
        visited = ['0'] * n
        memo = {}
        nums.sort(reverse = True)

        def backtracking(index, curr, count):
            currSum = sum(curr)
            visited_str = ''.join(visited)

            if all(element == '1' for element in visited) and currSum == target:
                return True
            if visited_str in memo:
                return memo[visited_str]

            if currSum == target:
                memo[visited_str] = backtracking(0, [], visited)
                return memo[visited_str]
            
            for i in range(index, n):
                if visited[i] == '0':
                    if nums[i] + currSum <= target:
                        visited[i] = '1'
                        curr.append(nums[i])
                        if backtracking(i + 1, curr, visited):
                            return True
                        curr.pop()
                        visited[i] = '0'
            
            return False
        
        return backtracking(0, [], visited)
        


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