Note that it is the k^th smallest element in the sorted order, not the k^th distinct element.

You must find a solution with a memory complexity better than O(n²).

Example 1:

Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8^th smallest number is 13

Example 2:

Input: matrix = [[-5]], k = 1
Output: -5

Constraints:

n == matrix.length == matrix[i].length
1 <= n <= 300
-10⁹ <= matrix[i][j] <= 10⁹
All the rows and columns of matrix are guaranteed to be sorted in non-decreasing order.
1 <= k <= n²

Follow up:

Could you solve the problem with a constant memory (i.e., O(1) memory complexity)?
Could you solve the problem in O(n) time complexity? The solution may be too advanced for an interview but you may find reading this paper fun.

23:T4ab,Given an `n x n` `matrix` where each of the rows and columns is sorted in ascending order, return _the_ `kth` _smallest element in the matrix_. Note that it is the `kth` smallest element **in the sorted order** , not the `kth` **distinct** element. You must find a solution with a memory complexity better than `O(n2)`. **Example 1:** **Input:** matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8 **Output

Kth Smallest Element in a Sorted Matrix

Problem Description

Problem Description

Answer Panel