<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 Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        def count_less_or_equal(val):
            count = 0
            n = len(matrix)
            i = n - 1
            j = 0 
            while i >= 0 and j < n:
                if matrix[i][j] <= val:
                    count += i + 1
                    j += 1
                else:
                    i -= 1
            return count


        left = matrix[0][0]
        right = matrix[-1][-1]

        while left <= right:
            mid = (left + right) // 2
            if count_less_or_equal(mid) >= k:
                right = mid - 1
            else:
                left = mid + 1
        return left
21000