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Formatted question description: https://leetcode.ca/all/2545.html
2545. Sort the Students by Their Kth Score
Description
There is a class with m
students and n
exams. You are given a 0indexed 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}
(0indexed) 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] <= 10^{5}
score
consists of distinct integers.0 <= k < n
Solutions
Solution 1: Sorting
We sort score in descending order based on the scores in the $k^{th}$ column, and then return it.
The time complexity is $O(m \times \log m)$, and the space complexity is $O(1)$. Here, $m$ is the number of rows in score.

class Solution { public int[][] sortTheStudents(int[][] score, int k) { Arrays.sort(score, (a, b) > b[k]  a[k]); return score; } }

class Solution { public: vector<vector<int>> sortTheStudents(vector<vector<int>>& score, int k) { sort(score.begin(), score.end(), [&](const auto& a, const auto& b) { return a[k] > b[k]; }); return score; } };

class Solution: def sortTheStudents(self, score: List[List[int]], k: int) > List[List[int]]: return sorted(score, key=lambda x: x[k])

func sortTheStudents(score [][]int, k int) [][]int { sort.Slice(score, func(i, j int) bool { return score[i][k] > score[j][k] }) return score }

function sortTheStudents(score: number[][], k: number): number[][] { return score.sort((a, b) => b[k]  a[k]); }

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 } }