<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>

import heapq

class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        matrixSize = len(matrix) * len(matrix)
        flattenedHeap = matrix[0]
        result = 0

        for i in range(1, len(matrix)):
            flattenedHeap.extend(matrix[i])
        
        heapq.heapify(flattenedHeap)

        while flattenedHeap and k > 0:
            result = heapq.heappop(flattenedHeap)
            k -= 1
        
        return result


21400