Counting the number of 1 bits in a bit string s can be accomplished in Java by first initializing an integer counter n to 0. Next, we implement a loop which iterates over each character c of s and when c is '1' we add 1 to n. When c is '0' we do nothing. Here is the implementation of that algorithm: private int countOnes(String s) { int n = 0; for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) == '1') { ++n; } } return n; This problem is also amenable to being solved using a recursive method rather than an iterative method (one that employs a loop). Your job is to select the correct pieces of code from those available and arrange them in proper order to implement a recursive method with the same method signature. private int countOnes(String s) { [LINE1] [LINE2] [LINE3] [LINE4] [LINE5] [LINE6] [LINE7] [LINE8] [LINE9]

Write a Java program to find the longest common prefix among a given array of strings. If there is no common prefix, return an empty string "".

def hailstone(n):  h.append(int(n))  if n == 1:    return h  while n != 1:    if n % 2 == 0:      return hailstone(int(n/2))    else:      return hailstone(int(3*n + 1))y=int(input("N: "))if y <= 0:    print("N must be positive!")else:    print("Hailstone sequences for 1 to " + str(y) + ":")    m=0;    i=1    for x in range(1, y+1):         n = i        h = []        h = hailstone(n)        print(h)         s=len(h)        if s > m:            m=s            flag=n        i=i+1    print("First longest sequence found in n =", + flag )     If i don't one the output have the square braket? How can I fix this code? Can you pleasw help me?

JAVA Language:  Transpose Rotate.  Question:  Modify Transpose.encode() so that it uses a rotation instead of a reversal. That is, a word like “hello” should be encoded as “ohell” with a rotation of one character. (Hint: use a loop to append the letters into a new string)         import java.util.*;            public class TestTranspose {        public static void main(String[] args) {          String plain = "this is the secret message";   // Here's the message           Transpose transpose = new Transpose();          String secret = transpose.encrypt(plain);          System.out.println("\n ********* Transpose Cipher Encryption *********");          System.out.println("PlainText: " + plain);    // Display the results          System.out.println("Encrypted: " + secret);          System.out.println("Decrypted: " + transpose.decrypt(secret));// Decrypt        }      }            abstract class Cipher {        public String encrypt(String s) {          StringBuffer result = new…

def hailstone(n):  h.append(int(n))  if n == 1:    return h  while n != 1:    if n % 2 == 0:      return hailstone(int(n/2))    else:      return hailstone(int(3*n + 1))y=int(input("N: "))if y <= 0:    print("N must be positive!")else:    print("Hailstone sequences for 1 to " + str(y) + ": ")    m=0;    i=1    for x in range(1, y+1):         n = i        h = []        h = hailstone(n)        print(*h)        print(",")        s=len(h)        if s > m:            m=s            flag=n        i=i+1    print("First longest sequence found in n =", + flag )   How can I fix this code if I want a comma between each number, not each line?

Convert the following java code to C++   //LabProgram.javaimport java.util.Scanner;public class LabProgram {public static void main(String[] args) {//defining a Scanner to read input from the userScanner input = new Scanner(System.in);//reading the value for Nint N = input.nextInt();//creating a 1xN matrixint[] m1 = new int[N];//creating an NxN matrixint[][] m2 = new int[N][N];//creating a 1xN matrix to store the resultint[] result = new int[N];//looping and reading N integers into m1for (int i = 0; i < N; i++) {m1[i] = input.nextInt();}//looping from 0 to N-1for (int i = 0; i < N; i++) {//looping from 0 to N-1for (int j = 0; j < N; j++) {//reading an integer and storing it into m2 at position i,jm2[i][j] = input.nextInt();}}//multiply m1 and m2, store result in result//looping from 0 to N-1for (int i = 0; i < N; i++) {//looping from 0 to N-1for (int j = 0; j < N; j++) {//multiplying value at index j in m1 with value at j,i in m2, adding to current value at index i// on…

JAVA: Write a recursive method named threeSum that accepts a list of integers and prints all combinations of three integers in the list that sum to 0. For example, if given the list [-1, 0, 1, 2, -1, -4], print the following lines of output: [-1, 0, 1] [-1, 2, -1] [0, 1, -1] You may print the lines of output in any order. The elements in each three-element sublist should appear in the same relative order that they appeared in the original list. Do not print duplicate lists; if the same exact sublist can be made in multiple ways, print it only once. If there are no combinations of three elements that sum to 0, print no output. The list passed to your method must be back to its original state at the end of the call. Either do not modify it, or if you modify it, fully undo your modifications before the method returns. Constraints: Do not declare any global variables. You can use any data structures you like, and your code can contain loops, but the overall algorithm must be recursive and…

I need help with creating a Java program described below:   Hamming distance. The Hamming distance between two-bit strings of length n is equal to the number of bits in which the two strings differ. Write a program that reads in an integer k and a bit string s from the command line, and prints all bit strings that have Hamming distance at most k from s.   For example, if k is 2 and s is 0000, then your program should print                           0011 0101 0110 1001 1010 1100   Hint: Choose k as the number of bits to flip in s.

Write the Java Program to Find the Elements that do Not have Duplicates. Given an array, print all the elements whose frequency is one, that is they do not have duplicates. Example: Array = [-1, -2, 3, 3, -2] Output = -1