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The arrays roll[] and marks[] store student roll numbers and their corresponding marks. The objective is to reorder the roll numbers such that they follow the descending order of marks, while maintaining the correct roll-marks association.

Example:

Input: roll[] = {101, 108, 103, 105}, marks[] = {70, 80, 40, 90}
Output: {105, 108, 101, 103}
Explanation: After sorting marks[] in decreasing order as {90, 80, 70, 40}, the corresponding roll[] values are rearranged to {105, 108, 101, 103} respectively.

Approach

Follow these steps to solve the problem efficiently:

Create a custom Student class to store marks and the corresponding roll number together
Traverse both arrays and store each marks–roll pair as a Student object in an ArrayList
Sort the list using Collections.sort() based on marks in decreasing order
Traverse the sorted list and print the roll numbers in the new order based on marks

Java

import java.util.*;

class Student {
    int marks;
    int roll;

    Student(int marks, int roll) {
        this.marks = marks;
        this.roll = roll;
    }
}

class GFG {

    static void sortRollByMarks(int[] roll, int[] marks) {
        int n = roll.length;

        ArrayList<Student> list = new ArrayList<>();

        // Create pairs of marks and roll numbers
        for (int i = 0; i < n; i++) {
            list.add(new Student(marks[i], roll[i]));
        }

        // Sort by marks in decreasing order
        Collections.sort(list, (a, b) -> b.marks - a.marks);

        // Print sorted roll numbers
        for (Student s : list) {
            System.out.print(s.roll + " ");
        }
    }

    public static void main(String[] args) {
        int[] roll = {101, 108, 103, 105};
        int[] marks = {70, 80, 40, 90};

        sortRollByMarks(roll, marks);
    }
}

Output

105 108 101 103

Time Complexity: O(N log N), where N is the number of students.
Space Complexity: O(N) due to the ArrayList storing the student objects.

Sort students by marks in Java

Approach

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