<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) {
        using t = pair <int, vector <int>>;
        auto comp = [] (const t & t1, const t &t2)
        {
            return t1.first > t2.first;
        };

        int n = mat [0].size ();
        priority_queue <t, vector <t>, decltype (comp)> q (comp);
        set <vector <int>> visited_idx;

        vector <int> idx0 (mat.size ());
        int sum {};
        for (const auto & r : mat) sum += r [0];

        visited_idx.insert (idx0);
        q.push ({sum, std::move (idx0)});
        int res;

        while (k--)
        {
            auto  sm = std::move (q.top ());
            q.pop ();
            res = sm.first;

            for (int i = 0; i < mat.size (); ++i)
            {
                auto & v = sm.second;
                int idx = v [i];
                if (idx != n - 1) {
                    auto v2 = v;
                    int s = sm.first - mat [i][idx] + mat [i][idx + 1];
                    v2 [i] = idx + 1;
                    if (!visited_idx.contains (v2))
                    {
                        visited_idx.insert (v2);
                        q.push ({s, std::move (v2)});
                    }
                    // v2 [i] += 1;

                }
            }
        }

        return res;
    }
};
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