<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:

vector<int> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        auto cmp = [&nums1, &nums2](pair<int,int> a, pair<int,int>b) {
            return nums1[a.first] + nums2[a.second] >  nums1[b.first] + nums2[b.second];
        };

        int n = nums1.size();
        int m = nums2.size();
        
        vector<int> result;
        
        if(n == 0 || m == 0) return {};
        
        priority_queue<pair<int,int>, vector<pair<int,int>>, decltype(cmp)> pq(cmp);
        pq.push({0, 0});
        
        while(k-- && !pq.empty()) {
            auto tmp = pq.top(); pq.pop();

            int i = tmp.first;
            int j = tmp.second;
            
            if (j + 1 < m)
                pq.push({i, j + 1});

            if (j == 0 && i + 1 < n)
                pq.push({i + 1, j});
            
            result.push_back(nums1[i] + nums2[j]);
        }

        return result;
    }
  
    int kthSmallest(vector<vector<int>>& mat, int k) {
        vector<int> nums1 = mat[0];
        int n =  mat.size();

        for(int i = 1; i < n; i++)
            nums1 = kSmallestPairs(nums1, mat[i], k);
        
        
        return nums1[k - 1];
    }
};
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