<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:
    int sum;
    vector<int> nums;
    int fullState;

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int total = 0;
        for (int i = 0; i < nums.size(); i++)
            total += nums[i];
        
        if (total % k != 0) return false;

        this->nums = nums;
        this->sum = total/k;
        this->fullState = (1 << nums.size()) - 1;

        vector<char> cache(65536, 0);
        return solve(k, sum, 0, cache) == 1;
    }

    char solve(int k, int s, int state, vector<char>& cache) {
        if (k == 0) {
            if (state != fullState) return 2;
            return 1;
        }

        for (int i = 0; i < nums.size(); i++) {
            int newS = s - nums[i];
            if ((state & (1 << i)) == false && newS >= 0) {
                if (newS == 0) {
                    k -= 1;
                    newS = sum;
                }

                int newState = (state | (1 << i));
                if (cache[newState] == 0) cache[newState] = solve(k, newS, newState, cache);
                if (cache[newState] == 1) return 1;
            }
        }

        return 2;
    }
};
20506