<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:
    vector<bool> visited;
    unordered_set<int> dp;
    bool Dfs(vector<int>& nums, int idx, int sum, int cur, int target, int k){
        if(dp.count(cur))
            return false;

        if(sum == 0 && k == 1)
            return true;

        if(sum == 0){
            return Dfs(nums, 0, target, cur, target, k-1);
        }

        if(idx == nums.size() || sum < nums[idx])
            return false;

        dp.insert(cur);
        for(int i=idx; i<nums.size();i++){
            //if(visited[i]) continue;
            int prev = cur;
            cur |= (1 << i);
            //visited[i] = true;
            if(Dfs(nums, i+1, sum-nums[i], cur, target, k))
                return true;
            cur = prev;
            //visited[i] = false;
        }

        return false;
    }

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if(sum % k)
            return false;
        sum = sum / k;
        sort(nums.begin(), nums.end());
        visited.resize(nums.size(), false);
        return Dfs(nums, 0, sum, 0, sum, k);
    }
};
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