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public class Test {
public static void main(String[] args) {
int[] x = {0, 1, 2, 3, 4, 5};
xMethod(x, 5);
}
public static void xMethod(int[] x, int length) {
for (int i = 0; i < length; i++)
System.out.print(" " + x[i]);
}
}
The
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The program displays 0 1 2 3 4 and then raises a runtime exception.
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The program displays 0 1 2 3 4.
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The program displays 0 1 2 3 4 5.
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- In Java trace the table for a while loop public class Main { public static void main(String[] args) { // Question 2 j = 30; x=5; for (int i=1;i<=2;i++) { while (j > 20) { j=j-x*i; ++x; }// end of while }// end of for j x i j<=2 j>20PROGRAM SIMULATION. Understand the given JAVA program and write the output. b. public class ChangeIt { public void doIt( int[] z ) { z[0] = 0; } } public class TestIt { public static void main ( String[] args ) { int[] myArray = {1, 2, 3, 4, 5} ; ChangeIt.doIt(myArray ); for (int j=0; j<myArray.length; j++ ) System.out.print( myArray[j] + " " ) ; } }Please help debug this: Debugging Exercise 4-3, Farrell, Joyce, Java Programming, 9th edition, Cengage Learning public class DebugBox { private int width; private int length; private int height; private FixDebugBox() { length = 1; width = 1; height = 1; } public DebugBox(int width, int length, int height) { width = width; length = length; height = height; } public void showData() { System.out.println("Width: " + width + " Length: " + length + " Height: " + height); } public double getVolume() { double vol = length - width - height; return vol; } }
- 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?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?Java - public static void test_a(int n) { System.out.println(n + " "); if (n>0) test_a(n-2); } What is printed by the call test_a(6)?
- In Java trace the table for the do while loop public class Main { public static void main(String[] args) { int x,j,y; /// Question 1 j = 30; x=5; do { j=j-x-2; --x; }while (j > 20); j x j>20Complete the sumOfDiagonals method in SumOfDiagonals.java to do the following public class SumOfDiagonals { public static void main(String[] args) { String[][] s1 = {{"2", "3", "4"}, {"5", "6", "7"}, {"8", "9", "10"}}; String[][] s2 = {{"1", "2", "3"}, {"5", "6", "7"}, {"9", "10", "11"}, {"13", "14", "15"}}; String[][] s3 = {{"1", "2", "3", "4"}, {"5", "6", "7", "8"}, {"9", "10", "11", "12"}, {"13", "14", "15", "16"}}; try { sumOfDiagonals(s1); sumOfDiagonals(s2); sumOfDiagonals(s3); } catch(NumberFormatException ex) { System.out.println("Not a valid integer"); } catch(IllegalArgumentException ex) { System.out.println("Array not a square"); } } public static void…Write a recursive Java method that takes a String as a parameter and returns true if the String is a palindrome. You may assume that the string only contains lower case characters (i.e. no numbers, spaces, punctuation, etc).
- Please add algorithm and comments to my program. Thank youpublic class For {public static void main(String[] args) {String days[] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday","Saturday", "Sunday"};for (int i = 0; i < days.length; i++) {// weekday =days[i] ;System.out.println(days[i]);}}}public class Soru32{ // Returns length of LCS for X[0..m-1], Y[0..n-1] static void lcs(String X, String Y, int m, int n) { int[][] L = new int[m+1][n+1]; // Following steps build L[m+1][n+1] in bottom up fashion. Note // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] for (int i=0; i<=m; i++) { for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (X.charAt(i-1) == Y.charAt(j-1)) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = Math.max(L[i-1][j], L[i][j-1]); } } // Following code is used to print LCS int index = L[m][n]; int temp = index; // Create a character array to store the lcs string char[] lcs = new char[index+1]; lcs[index] = '\u0000'; // Set the terminating character // Start from the right-most-bottom-most…public class Soru32{ // Returns length of LCS for X[0..m-1], Y[0..n-1] static void lcs(String X, String Y, int m, int n) { int[][] L = new int[m+1][n+1]; // Following steps build L[m+1][n+1] in bottom up fashion. Note // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] for (int i=0; i<=m; i++) { for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (X.charAt(i-1) == Y.charAt(j-1)) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = Math.max(L[i-1][j], L[i][j-1]); } } // Following code is used to print LCS int index = L[m][n]; int temp = index; // Create a character array to store the lcs string char[] lcs = new char[index+1]; lcs[index] = '\u0000'; // Set the terminating character // Start from the right-most-bottom-most…