<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, k):
        n = len(nums)
        # 排除一些基本情况
        if k > n:
            return False
        # nums数组的总和不能被均分, 直接返回False
        if sum(nums) % k != 0:
            return False

        target = sum(nums) // k
        # 在第一个子集中使用过的元素在第二个子集里就不能继续使用了
        # 所以必须初始化一个visited数组来记录
        visited = [False] * n
        memo = {}

        def backtracking(sum_, index, k):
            nonlocal visited, memo
            if k == 0:
                return True

            state = str(visited)
            
            if sum_ == target:
                ans = backtracking(0, 0, k - 1)
                memo[state] = ans

            if state in memo:
                return memo[state]

            for i in range(index, n):
                if visited[i] == True:
                    continue
                if sum_ + nums[i] > target:
                    continue

                sum_ += nums[i]
                visited[i] = True
                if backtracking(sum_, i + 1, k):
                    return True
                sum_ -= nums[i]
                visited[i] = False
            
            return False

        return backtracking(0, 0, k)
                

        
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