<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 {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int target = accumulate(nums.begin(), nums.end(), 0);

        if (target % k) { return false; }

        target /= k;
        sort(nums.begin(), nums.end());
        vector<char> cache(1 << nums.size(), -1);

        return checkPartition(nums, cache, 0, k, 0, target, target);
    }
private:
    int checkPartition(vector<int>& nums, vector<char>& cache, uint16_t used, char bucket, char from, int rest, int target) {
        if (bucket == 1) { return true; }
        if (cache[used] != -1) { return cache[used]; }

        for (int i = from; i < nums.size(); ++i) {
            if (used & 1 << i || nums[i] > rest) { continue; }

            if (rest == nums[i] && checkPartition(nums, cache, used | 1 << i, bucket-1, 0, target, target)) {
                return cache[used] = 1;
            } else if (checkPartition(nums, cache, used | 1 << i, bucket, i+1, rest-nums[i], target)) {
                return cache[used] = 1;
            }
        }

        return cache[used] = 0;
    }
};
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