<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 k > len(nums):
            return False

        total_sum = sum(nums)
        if total_sum % k != 0:
            return False
        target = total_sum // k
        memo = {}
        used = [0] * len(nums)

        def backtrack(index: int, bucket_sum: int, used: List[int], start: int) -> bool:
            if index == k:
                return True
            
            state = ''.join(map(str, used))
            if state in memo:
                return memo[state]
            
            is_found = False
            for i in range(start, len(nums)):
                if is_found:
                    break

                if used[i]:
                    # nums[i] is already in other bucket.
                    continue
                if nums[i] + bucket_sum > target:
                    continue
                
                used[i] = 1
                bucket_sum += nums[i]
                if bucket_sum == target:
                    is_found |= backtrack(index + 1, 0, used, 0)
                else:
                    is_found |= backtrack(index, bucket_sum, used, i+1)
                    
                used[i] = 0
                bucket_sum -= nums[i]

            state = ''.join(map(str, used))
            memo[state] = is_found
            return is_found
        
        return backtrack(0, 0, used, 0)
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