<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:
    unordered_map<string,bool> mp;
    bool sol(vector<int>& nums, int index, string &visited, int cnt, int k, int currSum, int target){
        int n=nums.size();
        if(mp.find(visited)!=mp.end()) return mp[visited];
        if(cnt == k-1) return true;
        if(currSum==target) return mp[visited] = sol(nums,0,  visited, cnt+1, k, 0, target);
        if(currSum > target) return false;

        for(int i=index; i<n; i++){
            if(visited[i]=='0'){
                visited[i]='1';
                if(sol(nums, i+1,  visited, cnt, k, currSum+nums[i], target))return true;
                visited[i]='0';
            }
        }
        return mp[visited]=false;
    }
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        string visited(nums.size(), '0');
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if(sum % k) return false;
        return sol(nums, 0, visited, 0, k, 0, sum/k);
    }
};
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