<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:
            return False
        nums.sort(reverse= True)
        target= sum(nums)/k
        if target < nums[0]:
            return False
        ans=dict()

        def backtrack(i, k, subsetSum, used):
            if k ==0:
                ans[used] = True
                return True
            if used in ans:
                return ans[used]
            if subsetSum == target:
                return backtrack(0, k-1, 0, used)
            for j in range(i, len(nums)):
                location = 1 << j
                if (used&location) > 0 or subsetSum + nums[j]>target:
                    continue
                used ^= location
                if backtrack(j+1, k, subsetSum+ nums[j], used):
                    ans[used]=True
                    used ^= location
                    return True
                used ^= location
            ans[used] = False
            return False

        return backtrack(0, k, 0, 0)

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