<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:
    using MinHeap = priority_queue<pair<int, vector<int>>, vector<pair<int, vector<int>>>, greater<>>;

    int kthSmallest(vector<vector<int>>& mat, int k) {
        MinHeap minHeap;
        int m = mat.size(), n = mat[0].size();

        int sum = 0;
        for (int i = 0; i < m; ++i) {
            sum += mat[i][0];
        }        

        set<vector<int>> visited;
        minHeap.push({ sum, vector<int>(m, 0) });

        while(--k) {
            auto cur = minHeap.top();
            minHeap.pop();

            int curSum = cur.first;
            auto cursor = cur.second;

            for (int i = 0; i < m; ++i) {
                if (cursor[i] + 1 < n) {
                    auto nextCursor = cursor;
                    nextCursor[i]++;

                    if (!visited.contains(nextCursor)) {
                        visited.insert(nextCursor);
                        int nextSum = curSum + mat[i][cursor[i]+1] - mat[i][cursor[i]];
                        minHeap.push({ nextSum, nextCursor });
                    }
                }
            }
        }

        return minHeap.top().first;
    }
};
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