LeetCode #378 — MEDIUM

Kth Smallest Element in a Sorted Matrix

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the k^th smallest element in the matrix.

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.

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[1,5,9],[10,11,13],[12,13,15]]
8

Example 2

[[-5]]
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        int left = matrix[0][0], right = matrix[n - 1][n - 1];
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (check(matrix, mid, k, n)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private boolean check(int[][] matrix, int mid, int k, int n) {
        int count = 0;
        int i = n - 1, j = 0;
        while (i >= 0 && j < n) {
            if (matrix[i][j] <= mid) {
                count += (i + 1);
                ++j;
            } else {
                --i;
            }
        }
        return count >= k;
    }
}

// Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
func kthSmallest(matrix [][]int, k int) int {
	n := len(matrix)
	left, right := matrix[0][0], matrix[n-1][n-1]
	for left < right {
		mid := (left + right) >> 1
		if check(matrix, mid, k, n) {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

func check(matrix [][]int, mid, k, n int) bool {
	count := 0
	i, j := n-1, 0
	for i >= 0 && j < n {
		if matrix[i][j] <= mid {
			count += (i + 1)
			j++
		} else {
			i--
		}
	}
	return count >= k
}

# Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        def check(matrix, mid, k, n):
            count = 0
            i, j = n - 1, 0
            while i >= 0 and j < n:
                if matrix[i][j] <= mid:
                    count += i + 1
                    j += 1
                else:
                    i -= 1
            return count >= k

        n = len(matrix)
        left, right = matrix[0][0], matrix[n - 1][n - 1]
        while left < right:
            mid = (left + right) >> 1
            if check(matrix, mid, k, n):
                right = mid
            else:
                left = mid + 1
        return left

// Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
struct Solution;
use std::collections::BinaryHeap;

impl Solution {
    fn kth_smallest(matrix: Vec<Vec<i32>>, k: i32) -> i32 {
        let mut pq: BinaryHeap<i32> = BinaryHeap::new();
        for row in matrix {
            for x in row {
                pq.push(x);
                if pq.len() > k as usize {
                    pq.pop();
                }
            }
        }
        pq.pop().unwrap()
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[1, 5, 9], [10, 11, 13], [12, 13, 15]];
    let k = 8;
    let res = 13;
    assert_eq!(Solution::kth_smallest(matrix, k), res);
    let matrix = vec_vec_i32![[1, 2], [1, 3]];
    let k = 3;
    let res = 2;
    assert_eq!(Solution::kth_smallest(matrix, k), res);
}

// Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #378: Kth Smallest Element in a Sorted Matrix
// class Solution {
//     public int kthSmallest(int[][] matrix, int k) {
//         int n = matrix.length;
//         int left = matrix[0][0], right = matrix[n - 1][n - 1];
//         while (left < right) {
//             int mid = (left + right) >>> 1;
//             if (check(matrix, mid, k, n)) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left;
//     }
// 
//     private boolean check(int[][] matrix, int mid, int k, int n) {
//         int count = 0;
//         int i = n - 1, j = 0;
//         while (i >= 0 && j < n) {
//             if (matrix[i][j] <= mid) {
//                 count += (i + 1);
//                 ++j;
//             } else {
//                 --i;
//             }
//         }
//         return count >= k;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.