<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:
        # memoization solution
        if sum(nums) % k != 0:
            return False
        target_sum = sum(nums) / k
        visited = set()

        def backtrack(count, cur_sum, index, taken):
            if "".join(taken) in visited:
                return False
            if count == k-1:
                return True
            if cur_sum == target_sum:
                return backtrack(count+1, 0, 0, taken)
            if cur_sum > target_sum:
                return False
            for i in range(index, len(nums)):
                if taken[i] == '0':
                    taken[i] = '1'
                    if backtrack(count, cur_sum+nums[i], i+1, taken):
                        return True
                    taken[i] = '0'
            visited.add("".join(taken))
            return False

        taken = ['0'] * len(nums)
        return backtrack(0, 0, 0, taken)

        
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