<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 == 0 or n < k:
            return False

        if k <= 1:
            return True

        if sum(nums) % k != 0:
            return False

        target_sum = sum(nums) / k
        nums.sort()
        nums.reverse()

        if nums[-1] > target_sum:
            return False

        n = len(nums)

        def hash_param(i: int, group_sums: List[int]) -> str:
            group_sums_joined = '_'.join([str(i) for i in sorted(group_sums)])
            return f"{i}__{group_sums_joined}"

        memo = set() # failed sequence

        def dfs(i: int, group_sums: List[int]) -> bool:
            if i == n:
                return True

            if hash_param(i, group_sums) in memo:
                return False

            for g in range(k): #to join grp k
                if group_sums[g] + nums[i] > target_sum:
                    continue

                group_sums[g] += nums[i]

                if dfs(i+1, group_sums):
                    return True
                else:
                    group_sums[g] -= nums[i]

            memo.add(hash_param(i, group_sums))
            return False

        return dfs(0, [0] * k)


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