<p>Given an <code>n x n</code> <code>matrix</code> where each of the rows and columns is sorted in ascending order, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest element in the matrix</em>.</p>

<p>Note that it is the <code>k<sup>th</sup></code> smallest element <strong>in the sorted order</strong>, not the <code>k<sup>th</sup></code> <strong>distinct</strong> element.</p>

<p>You must find a solution with a memory complexity better than <code>O(n<sup>2</sup>)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
<strong>Output:</strong> 13
<strong>Explanation:</strong> The elements in the matrix are [1,5,9,10,11,12,13,<u><strong>13</strong></u>,15], and the 8<sup>th</sup> smallest number is 13
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[-5]], k = 1
<strong>Output:</strong> -5
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>-10<sup>9</sup> &lt;= matrix[i][j] &lt;= 10<sup>9</sup></code></li>
	<li>All the rows and columns of <code>matrix</code> are <strong>guaranteed</strong> to be sorted in <strong>non-decreasing order</strong>.</li>
	<li><code>1 &lt;= k &lt;= n<sup>2</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Could you solve the problem with a constant memory (i.e., <code>O(1)</code> memory complexity)?</li>
	<li>Could you solve the problem in <code>O(n)</code> time complexity? The solution may be too advanced for an interview but you may find reading <a href="http://www.cse.yorku.ca/~andy/pubs/X+Y.pdf" target="_blank">this paper</a> fun.</li>
</ul>

class MinHeap {
public:
    bool operator()(const tuple<int, int, int>& a,
                    const tuple<int, int, int>& b) {
        return get<0>(a) > get<0>(b);
    }
};

class Solution {
public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int rowCount = matrix.size();
        priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>,
                       MinHeap>
            minHeap;

        for (int i = 0; i < min(rowCount, k); ++i) {
            minHeap.push(make_tuple(matrix[i][0], i, 0));
        }

        int numbersChecked = 0, smallestElement = 0;

        while (!minHeap.empty()) {
            auto [value, rowIndex, colIndex] = minHeap.top();
            minHeap.pop();
            smallestElement = value;
            numbersChecked++;

            if (numbersChecked == k)
                break;

            if (colIndex + 1 < matrix[rowIndex].size()) {
                minHeap.push(make_tuple(matrix[rowIndex][colIndex + 1],
                                        rowIndex, colIndex + 1));
            }
        }

        return smallestElement;
    }
};
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