<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:
    map<int,bool> vis;
    bool backtrack(int i, int curr_sum, int k, int target, vector<int>& nums)
    {
        if(k==0)
            return true;
        if(curr_sum==target)
            return backtrack(0,0,k-1,target,nums);
        for(int j=i;j<nums.size();j++)
        {
            if(vis[j] || curr_sum+nums[j]>target)
                continue;
            if(j>0 && !vis[j-1] && nums[j]==nums[j-1])
                continue;
            vis[j] = true;
            if(backtrack(j+1,curr_sum+nums[j],k,target,nums))
                return true;
            vis[j] = false;
        }
        return false;

    }
    bool canPartitionKSubsets(vector<int>& nums, int k) 
    {
        int s = accumulate(nums.begin(),nums.end(),0);
        if(s%k!=0)
            return false;
        sort(nums.begin(),nums.end());
        int target = s/k;
        return backtrack(0,0,k,target,nums);
    }
};
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