The language used is Java /** * * Merging k sorted arrays into a single sorted array. * Your algorithm must be stable. * * You are NOT allowed to call the constructor of class Student, except i


The language used is Java /**  *   * Merging k sorted arrays into a single sorted array.  * Your algorithm must be stable.  *   * You are NOT allowed to call the constructor of class Student, except in <em>main</em>.  */ public class StableMerger {     /**      *       * No modification to class Student is allowed.      * If you change anything

This question hasn’t been solved yet

The language used is Java/** * * Merging k sorted arrays into a single sorted array. * Your algorithm must be stable. * * You are NOT allowed to call the constructor of class Student, except in <em>main</em>. */public class StableMerger {    /**     *     * No modification to class Student is allowed.     * If you change anything here, your work will not be graded.     *     */    static class Student {        String name;        double grade;        public Student(String n, double g) {            name = n; grade = g;        }        public String toString() {            return “(” + name + “, ” + grade + “)”;        }    } 

    /**     * Running time: O(   ). (n is the total number of students.)        */    public static Student[] merge(Student[][] a) {        return null;    } 

    public static void main(String[] args) {        String[][] names = {{“Gloria Tang”, “Andy Lau”, “McDull”},                {“Eason Chan”, “Denise Ho”, “Jennifer Chan”, “Joey Yung”, “Kay Tse”, “Jacky Cheung”, “Anita Mui”},                {“Winnie”, “Mickey”, “Teddy”, “Peppa”}};        double[][] grades = {{60, 60, 60}, {40, 60, 70, 80, 90, 95, 100}, {0, 90, 95, 100}};        Student[][] a = new Student[names.length][];        for (int i = 0; i < names.length; i++) {            a[i] = new Student[names[i].length];            for (int j = 0; j < names[i].length; j++)                  a[i][j]= new Student(names[i][j], grades[i][j]);        }        System.out.println(Arrays.toString(merge(a)));    }}