<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>

class Solution {
public:
    int kthSmallest(vector<vector<int>>& mat, int k) {
        // imagine we start with 
        // we start with an index and in each step do a for loop
        // now given a sum, can we calculate number of 
        // at each step, we can move in the following way
        //

        priority_queue< pair<int,vector<int>> > pq;

        int sum = 0;
        int n = mat.size();
        int m = mat[0].size();
        for(int i = 0 ; i<n ; i++)
            sum+= mat[i][0];
        
        vector<int> tmp (n , 0);

        pq.push({-sum , tmp});

        set<vector<int>> vis;

        for(int i = 1 ;i<k ;i++)
        {
            int sum = -pq.top().first;
            vector<int> tmp = pq.top().second;

            pq.pop();

            // cout<<sum<<" "<<tmp[0]<<" "<<tmp[1]<<endl;

            for(int j = 0 ;j<n ; j++)
            {
                vector<int> arr = tmp;

                if( (arr[j]+1) < m )
                {
                    arr[j]++;
                                // out<<arr[j]<<" ";
                                        int finsum = sum - mat[j][arr[j]-1] + mat[j][arr[j]];

                    // cout<<" "<<finsum<<" "<<arr[0]<<" "<<arr[1]<<endl;
                    if(vis.count(arr) == 0 )
                    {
                        vis.insert(arr);
                        pq.push({ -finsum, arr});

                    }
                    arr[j]--;
                }
            }
            // cout<<endl;
        }
        return -pq.top().first;  
    }
};
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