<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>

//Jai Shree Ram
//Jai Maa Durga

class Solution {
public:
    //For memoization need k=17, mask=(1e15)
    // curr_sum to redundandt hai kyuki mask curr_sum store kar lega

    int dp[17][1<<16];
    int solve(vector<int>&nums,int k,int curr_sum,int mask,int required_val){
        int n=nums.size();
        if(k<=0)return 1;

        if(dp[k][mask]!=-1)return dp[k][mask];
        if(curr_sum==required_val)return solve(nums,k-1,0,mask,required_val);
        else if(curr_sum>required_val)return 0;

        int ans=0;
        for(int i=0;i<n;i++){
            if(mask &(1<<i) || curr_sum+nums[i]>required_val)continue;
            ans=ans or solve(nums,k,curr_sum+nums[i],mask |(1<<i),required_val);
        }
        return dp[k][mask]=ans;
        
    }
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int sum=0;
        for(auto it:nums)sum+=it;
        if(sum%k!=0)return false;
        memset(dp,-1,sizeof(dp));
        return solve(nums,k,0,0,sum/k);
    }
};
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