<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)
    {
        sort(nums.rbegin(), nums.rend());
        int n = nums.size();
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if (sum % k != 0) return false;
        
        sum = sum / k;
        set<vector<int>> visited;
        queue<vector<int>> qu;
        int j = 0;
        qu.emplace(vector<int>(k, 0));
        
        while (!qu.empty() && j < n)
        {
            int qsize = qu.size();
            
            while(qsize--)
            {
                auto current = qu.front(); qu.pop();
                for (int i = 0; i < k; i++)
                {
                    auto next = current;
                    next[i] += nums[j];
                    if (next[i] > sum) continue;
                    sort(next.begin(), next.end());
                    if (visited.count(next) == 0)
                    {
                        visited.insert(next);
                        qu.emplace(next);
                    }
                }
            }
            j++;
        }
        
        while (!qu.empty())
        {
            auto items = qu.front(); qu.pop();
            bool flag = true;
            for (auto item : items)
            {
                if (item != sum)
                {
                    flag = false;
                    break;
                }
            }
            if (flag) return true;
        }
        
        return false;
    }
};
30619