<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 tar;
int dp[1000001];
bool ritik(vector<int>&nums,int mask,int k,int sum)
{
    if(mask==(1<<nums.size())-1) return false;
     if(k==1) return true;
     
     if(sum==tar) return  ritik(nums,mask,k-1,0);
     if(dp[mask]!=-1) return dp[mask];

        bool ans=false;

      for(int i=0;i<nums.size();i++)
      {
        if(mask&(1<<i)||sum+nums[i]>tar) continue;

        
        ans=ans | ritik(nums,mask|(1<<i),k,sum+nums[i]);
       

      }


      return dp[mask]= ans;
}
    bool canPartitionKSubsets(vector<int>& nums, int k) {

        int n=nums.size();
        int sum=0;
        for(auto it:nums)
        sum+=it;

       memset(dp,-1,sizeof(dp));
        tar=sum/k;
        if(sum%k!=0) return false;
        return ritik(nums,0,k,0);


        
    }
};
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