<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>>& matrix, int k) {
        int n = matrix.size();
        int m = matrix[0].size();
        long long int sum = 0;
        for (int i = 0; i<n; i++){
            sum += matrix[i][0];
        }
        vector<int>arr(n, 0);
        priority_queue<pair<long long int,vector<int>>, vector<pair<long long int,vector<int>>>, greater<pair<long long int,vector<int>>>>pq;
        pq.push({sum, arr});
        set<vector<int>> st;
        st.insert(arr);
        long long ans = 0;
        while(k--){
            auto x = pq.top();
            pq.pop();
            ans = x.first;
            vector<int>vis = x.second;
            for (int i = 0; i<n; i++){
                long long newS = x.first;
                if (vis[i] + 1 < m){
                    vis[i]++;
                    if (st.find(vis) == st.end()){
                        newS -= matrix[i][vis[i]-1];
                        newS += matrix[i][vis[i]];
                        pq.push({newS, vis});
                        st.insert(vis);
                    }
                    vis[i]--;
                }
            }
        }
        return ans;
    }
};
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