<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) {
        int m = mat.size(), n = mat[0].size();
        int left = m, right = m * 5000, ans = -1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            int cnt = countArraysHaveSumLessOrEqual(mat, m, n, mid, 0, 0, k);
            if (cnt >= k) {
                ans = mid;
                right = mid-1 ;
            } else left = mid + 1;
        }
        return ans;
    }
    int countArraysHaveSumLessOrEqual(vector<vector<int>>& mat, int m, int n, int targetSum, int r, int sum, int k) {
        if (sum > targetSum) return 0;
        if (r == m) return 1; // found a valid array with sum <= targetSum
        int ans = 0;
        for (int c = 0; c < n; ++c) {
            int cnt = countArraysHaveSumLessOrEqual(mat, m, n, targetSum, r + 1, sum + mat[r][c], k );
            if (cnt == 0) break; // prune, the array (which contains mat[r][c]) has sum > targetSum -> No need to process anymore
            ans += cnt;
            if (ans >=k) return ans;; // Important prune, since count > k -> No need to process anymore
        }
        return ans;
    }
};
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