<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 solve(vector<int>& nums,int k, int t, int curr,int currsum,int& vis,unordered_map<int,bool>& mp){
        if(currsum == t) {
            curr++;
            currsum = 0;
        }
        else if(currsum > t) return false;
        if(curr == k) return mp[vis] = true;
        if(mp.count(vis)) return mp[vis];
        for(int i=0;i<nums.size();i++){
            if( ( (vis>>i) & 1 ) == 0 ){
                vis = vis ^ (1<<i);
                if(solve(nums,k,t,curr,currsum+nums[i],vis,mp)) return mp[vis] =  true;
                vis = vis ^ (1<<i);
            }
        }
        return mp[vis] = false;
    }
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int sum = accumulate(nums.begin(),nums.end(),0);
        if(sum%k!=0) return false;
        int vis = 0, t = sum/k;
        unordered_map<int,bool> mp;
        sort(nums.rbegin(), nums.rend());
        if (nums[0] > t) return false;
        return solve(nums,k,t,0,0,vis,mp);
    }
};
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