<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 canPartitionKSubsets(vector<int>& nums, int k) {
        int n=nums.size();
        int mask=0;
        int sum=accumulate(nums.begin(),nums.end(),0);
        if(sum%k!=0){
            return false;
        }
        int target=sum/k;
        vector<int>dp((1 << n)+1,-1);
        function<bool(int,int,int)> rec = [&](int mask,int k,int curSum) ->bool {
            if(k==0){
                return true;
            }
            if(curSum<0){
                return false;
            }
            if(curSum==0){
                return rec(mask,k-1,target);
            }
            if(dp[mask]!=-1){
                return dp[mask];
            }
            for(int i=0;i<n;i++){
                if(mask & (1 << i)) continue;
                int newMask=mask | (1 << i);
                if(rec(newMask,k,curSum-nums[i])){
                    return dp[mask]=true;
                }
            }
            return dp[mask]=false;
        };
        return rec(mask,k,target);
    }
};
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