<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 {
    vector<int> helper(vector<vector<int>> &mat, vector<int> init, int i) {
        int n = mat.back().size();
        if (init[i] + 1 < n) {
            init[0] += mat[i - 1][init[i] + 1] - mat[i - 1][init[i]];
            init[i]++;
        }
        else {
            init[0] = INT_MAX;
        }
        return init;
    }
public:
    int kthSmallest(vector<vector<int>>& mat, int k) {
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        int sum = 0;
        int m = mat.size(), n = mat.back().size();
        for (int i = 0; i < m; ++i) {
            sum += mat[i][0];
        }
        vector<int> init(m + 1);
        init[0] = sum;
        pq.push(init);
        set<vector<int>> st;
        int result = pq.top()[0];
        vector<int> tmp;
        for (int i = 1; i < pq.top().size(); ++i) {
            tmp.push_back(pq.top()[i]);
        }
        st.insert(tmp);
        while (--k) {
            init = pq.top();
            pq.pop();
            for (int i = 1; i <= m; ++i) {
                vector<int> next = helper(mat, init, i);
                if (next[0] != INT_MAX) {
                    tmp = {};
                    for (int i = 1; i < next.size(); ++i) {
                        tmp.push_back(next[i]);
                    }
                    if (st.insert(tmp).second)
                        pq.push(next);

                }
            }
            result = pq.top()[0];
            //cout << result << ' ';
        }
        return result;
    }
};
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