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- Edit this code to be Allow to input a double number The code is //AssignmentMarks.java import java.util.Scanner; public class AssignmentMarks { public static void main(String[] args) { // defining number of students and assignments final int students_count = 5; final int assignments_count = 10; // initializing a 2d integer array int scores[][] = new int[students_count][assignments_count]; // scanner for reading input from keyboard Scanner sc = new Scanner(System.in); // looping through each index from 0 to students_count-1 for (int i = 0; i < students_count; i++) { // each row represent marks of each student. so asking user to enter // assignments_count marks for current student (one for each // assignment) System.out.print("Enter…Method Details: public static java.lang.String getArrayString(int[] array, char separator) Return a string where each array entry (except the last one) is followed by the specified separator. An empty string will be return if the array has no elements. Parameters: array - separator - Returns: stringThrows:java.lang.IllegalArgumentException - When a null array parameter is provided#include int main() { int a[5]; for (int i = = a[i] 1; for (int j = 0; j = a) { } = int n *pa; while (n >= 1) { n = n / 2; printf("%d *pa); pa-- ; 2 1; } Q. What is the output?
- public class ArrayMerge { static void arrayMerge(int a[], int b[]){ int c[] = new int[50]; int k = 0; for(int i = 0; i < a.length; i++) { c[k++] = a[i]; } for(int i = 0; i < b.length; i++) { c[k++] = b[i]; } for(int i = 0; i < k; i++) { System.out.println(c[i]); } } public static void main(String[] args) { int a[] = { 18, 36, 41, 78, 56 }; int b[] = { 10, 37, 44 }; arrayMerge(a,b); } } Calculate the algorithm step number and algorithm time complexity of the above program?URGENT JAVA Write a Java method that takes two 2 dimensional int arrays (a and b) and a 2 dimensional boolean array (c) where all three arrays have the same size. Your method should return an array (result) such that, if a partic- ular element in c is true, then the corresponding (same indexed) element in result must be the multiplication of the the corresponding elements of a and b. If the element in c is false, then the the corresponding element in result must be the negative of multiplication of the the corresponding element in a and b. For example ifa = {{1,2,3},{4,5}},b = {{1,2,1},{0,2}} andc = {{true,false,true},{false,true}}, then the result should be result={{1, -4, 3}{0, 10}}.Java Programming : At a fun fair, a street vendor is selling different colours of balloons. He sells N number of different colours of balloons (B[]). The task is to find the colour (odd) of the balloon which is present odd number of times in the bunch of balloons. Note: If there is more than one colour which is odd in number, then the first colour in the array which is present odd number of times is displayed. The colours of the balloons can all be either upper case or lower case in the array. If all the inputs are even in number, display the message 'All are even'. Example 1: 7 -> Value of N [r,g,b,b,g,y,y] -> B[] Elements B[0] to B[N-1], where each input element is separated by new line. Output: r -> [r,g,b,b,g,y.y] →> 'r' colour balloon is present odd number of times in the bunch. Explanation: From the input array above: r: 1 balloon g: 2 balloons b: 2 balloons y: 2 balloons Hence, r is only the balloon which is odd in number.
- public class ArraySection { static void arraySection(int a[], int b[]) { int k = 0; int [] c = new int[a.length]; for(int i = 0; i < a.length; i++) { for(int j = 0; j < b.length; j++) if(a[i] == b[j]) c[k++] = a[i]; } for(int i = 0; i < k; i++) System.out.println(c[i]); System.out.println(); } public static void main(String[] args) { int a[] = { 1, 2, 3, 4, 5 }; int b[] = { 0, 2, 4,5 }; arraySection(a,b); }} Calculate the algorithm step number and algorithm time complexity of the above program?#include #include main () { int array [3] [2] [2], value=0, i, j, k; int *p= (int *) array; } for (i = 0; i < 3; i++) { for(j= } = 0; j < 2; j++) { for (k = 0; k < 2; k++) { array[i][j] [k] = value; value = value + 1; } } for (i = 0; i < 12; i++) printf("%d ", p[i]); printf("\n"); printf ("Examine memory now.\n");using namespace std; int main() int i, m=0,n=4; float arr[100] (55,66,88,1); for(i=0; ipublic static void main(String[] args) { int[] funky = {1, -5, 3, -5, 6, 6, 6, 7, -5};int expectedFunkyRange = 5; // |6 - 1| = 5System.out.println("Expected funkyRange of " + Arrays.toString(funky) + " to be " + expectedFunkyRange + " and got " + funkyRange(funky));int[] funkyTwo = {6, 6, 4, 4, 4, 2, 2, 8, 4, 1, 1};int expectedFunkyRangeTwo = 4; // |4 - 8| = 4System.out.println("Expected funkyRange of " + Arrays.toString(funkyTwo) + " to be " + expectedFunkyRangeTwo + " and got " + funkyRange(funkyTwo)); Write a static method: public static int WeirdRange(int[] a) that that calculates the difference between the mode and "anti-mode" of an array of integers. This difference should be a positive number (you can use Math.abs to calculate an absolute value). If the array is empty or null, return -1. If the array has only one thing in it, return 0 (since the mode and anti-mode are the same). Definitions: The mode of an array is the element that occurs the most frequently.…public class FindArrayDifference { static void arrayDifference(int a[], int b[]) { int k = 0; int [] c = new int[a.length]; for(int i=0; i < a.length; i++) { int j; for(j = 0; j < b.length; j++) if(a[i] == b[j]) break; if(j == b.length) c[k++] = a[i]; } for(int j = 0; j < k; j++) System.out.println(c[j]); } public static void main(String[] args) { int a[] = {1,2,3,7,8,15,26}; int b[] = {1,2,3,15,4,8,6}; arrayDifference(a,b); } } Calculate the algorithm step number and algorithm time complexity of the above program?b) int [] num = new int [6]; for (int i = 1; i 2) { num [i] = 2 * num[i]-num[i-1]; System.out.println( "num [" + i + ") =" + num [i]);SEE MORE QUESTIONS