LeetCode #2545 — MEDIUM

Sort the Students by Their Kth Score

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

The Problem

Problem Statement

There is a class with m students and n exams. You are given a 0-indexed m x n integer matrix score, where each row represents one student and score[i][j] denotes the score the i^th student got in the j^th exam. The matrix score contains distinct integers only.

You are also given an integer k. Sort the students (i.e., the rows of the matrix) by their scores in the k^th (0-indexed) exam from the highest to the lowest.

Return the matrix after sorting it.

Example 1:

Input: score = [[10,6,9,1],[7,5,11,2],[4,8,3,15]], k = 2
Output: [[7,5,11,2],[10,6,9,1],[4,8,3,15]]
Explanation: In the above diagram, S denotes the student, while E denotes the exam.
- The student with index 1 scored 11 in exam 2, which is the highest score, so they got first place.
- The student with index 0 scored 9 in exam 2, which is the second highest score, so they got second place.
- The student with index 2 scored 3 in exam 2, which is the lowest score, so they got third place.

Example 2:

Input: score = [[3,4],[5,6]], k = 0
Output: [[5,6],[3,4]]
Explanation: In the above diagram, S denotes the student, while E denotes the exam.
- The student with index 1 scored 5 in exam 0, which is the highest score, so they got first place.
- The student with index 0 scored 3 in exam 0, which is the lowest score, so they got second place.

Constraints:

m == score.length
n == score[i].length
1 <= m, n <= 250
1 <= score[i][j] <= 10⁵
score consists of distinct integers.
0 <= k < n

Step 01

Brute Force Baseline

Problem summary: There is a class with m students and n exams. You are given a 0-indexed m x n integer matrix score, where each row represents one student and score[i][j] denotes the score the ith student got in the jth exam. The matrix score contains distinct integers only. You are also given an integer k. Sort the students (i.e., the rows of the matrix) by their scores in the kth (0-indexed) exam from the highest to the lowest. Return the matrix after sorting it.

Baseline thinking

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

Pattern signal: Array

Example 1

[[10,6,9,1],[7,5,11,2],[4,8,3,15]]
2

Example 2

[[3,4],[5,6]]
0

Core Insight

What unlocks the optimal approach

Find the row with the highest score in the kth exam and swap it with the first row.
After fixing the first row, perform the same operation for the rest of the rows, and the matrix's rows will get sorted one by one.

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 #2545: Sort the Students by Their Kth Score
class Solution {
    public int[][] sortTheStudents(int[][] score, int k) {
        Arrays.sort(score, (a, b) -> b[k] - a[k]);
        return score;
    }
}

// Accepted solution for LeetCode #2545: Sort the Students by Their Kth Score
func sortTheStudents(score [][]int, k int) [][]int {
	sort.Slice(score, func(i, j int) bool { return score[i][k] > score[j][k] })
	return score
}

# Accepted solution for LeetCode #2545: Sort the Students by Their Kth Score
class Solution:
    def sortTheStudents(self, score: List[List[int]], k: int) -> List[List[int]]:
        return sorted(score, key=lambda x: -x[k])

// Accepted solution for LeetCode #2545: Sort the Students by Their Kth Score
impl Solution {
    pub fn sort_the_students(mut score: Vec<Vec<i32>>, k: i32) -> Vec<Vec<i32>> {
        let k = k as usize;
        score.sort_by(|a, b| b[k].cmp(&a[k]));
        score
    }
}

// Accepted solution for LeetCode #2545: Sort the Students by Their Kth Score
function sortTheStudents(score: number[][], k: number): number[][] {
    return score.sort((a, b) => b[k] - a[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(m × log m)

Space

O(log m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.