<p>You are given an <code>m x n</code> matrix <code>mat</code> that has its rows sorted in non-decreasing order and an integer <code>k</code>.</p>

<p>You are allowed to choose <strong>exactly one element</strong> from each row to form an array.</p>

<p>Return <em>the </em><code>k<sup>th</sup></code><em> smallest array sum among all possible arrays</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 9
<strong>Output:</strong> 17
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
<strong>Output:</strong> 9
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat.length[i]</code></li>
	<li><code>1 &lt;= m, n &lt;= 40</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 5000</code></li>
	<li><code>1 &lt;= k &lt;= min(200, n<sup>m</sup>)</code></li>
	<li><code>mat[i]</code> is a non-decreasing array.</li>
</ul>

struct ls
{
int sum;
vector<int> ind;
};

struct comp
{
    bool operator () (ls l1,ls l2)
    {
        return l1.sum>l2.sum;  
    }
};

class Solution {
public:
    int kthSmallest(vector<vector<int>>& mat, int k) {
      set<vector<int> > st;
      int n= mat.size();
      int m= mat[0].size();
      int s=0;
      vector<int> temp(n,0);
        for(int i=0; i<n; i++)
      {
          s+=mat[i][0];
      }
      ls a;
      a.sum=s;
      a.ind=temp;
      priority_queue<ls,vector<ls>,comp> pq;
      st.insert(a.ind);
      pq.push(a);
        
      while(k--)
      {
       a=pq.top();
       pq.pop();
       for(int i=0; i<n; i++)
       {
         if(a.ind[i]+1<m)
         {
             ls t;
             t.ind=a.ind;
             t.ind[i]++;
             if(st.count(t.ind))
                 continue;                 
             t.sum=a.sum-mat[i][a.ind[i]]+mat[i][a.ind[i]+1];  
             pq.push(t);
             st.insert(t.ind);
         }
       }
      }
    return a.sum;
    }
};
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