<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 fixed;
    bool canPartitionKSubsets(vector<int>& nums, int k) {
       int target=0;
       for(auto n:nums){
            target+=n;
       }
       if(target%k!=0){
        return false;
       }
       target/=k;
       unordered_map<int,unordered_map<int,unordered_map<int,int>>> cache;
       fixed=target;
       return solve(nums,k,target,0,cache);
    }

    bool solve(vector<int>& nums,int k,int target,int visited,unordered_map<int,unordered_map<int,unordered_map<int,int>>>& cache){
        if(k==0 && visited==(1<<nums.size())-1){
            return true;
        }

        if(target<0){
            return false;
        }

        if(cache.find(k)!=cache.end() && cache[k].find(target)!=cache[k].end() && cache[k][target].find(visited)!=cache[k][target].end()){
            return cache[k][target][visited];
        }

        bool ans=false;
        for(int i=0;i<nums.size();i++){
            if(visited>>i & 1){
                continue;
            }

            visited=visited|(1<<i);

            if(target-nums[i]==0){
                ans=ans||solve(nums,k-1,fixed,visited,cache);
            } else {
                ans=ans||solve(nums,k,target-nums[i],visited,cache);
            }

            visited=visited&(~(1<<i));

        }
        return cache[k][target][visited]=ans;
    }
};

/*

4   3   2   3   5   2   1
*   *   *   *   *   *   *

k=0


20/4 = 5
k=4

*/
46799