public static void doubleArray(int[] a){ int[] newA - new int[a.length]; for (int i = 0; i < a.length; i++){ newA[i] - 2*a[i]; } a - newA; } // doubleArray public static void main(String[] args){ int[] a - new int[]{1,2,3,4,5,6}; doubleArray(a); for(int i - 0; i < a.length; i++){ System.out.print(a[i]); }
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Give the time complexity of the doubleArray() method provided. Briefly explain.
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- 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 getDup accepts an array and returns an int. If there is any element is duplicated, the method counts how many of those present in the array. Example : getDup(["1","2","aa"',"1"]) returns:2 ("1" is duplicated and there are two "1"s so return is 2) class Main { public static int getDup(String[] r) { } public static void main(String[] args) { Scanner in = new Scanner(System.in); String [] strs = new String[in.nextInt()]; for(int i=0; i < strs.length; i++) { strs[i] = in.next(); } System.out.println(getDup(strs)); } } hint: you will need a nested array the will run on the array we get from the method argument. inside the nested loop you will need to check if the main array element is equal to the currentWrite an application that displays the strings in the provided array alphabetically in ascending order. import java.util.*; public class StringSort { public static void main(String[] args) { String[] values = {"mouse", "dog", "cat", "horse", "cow", "moose", "tiger", "lion", "elephant", "bird", "hamster", "guinea pig", "leopard", "aardvark", "hummingbird"}; } }
- Complete the private combinations method. Modify the code so that it is possible to compute combinations(“ABCD”,2) with fewer than 20 calls and combinations(“ABCDE”,3) with fewer than 30 calls. import static org.junit.Assert.*; import java.util.ArrayList; import org.junit.Test; /** * Recursive computation of all combinations of k characters from a string s. * */public class Combinations { private static int recusiveCalls = 0; /** * Generate all combinations of the characters in the string s with length * k. * * @param s * a string. * @param k * the length of the combinations * @return a list of all of the combinations of the strings in s. */ public static ArrayList<String> combinations(String s, int k) { return combinations("", s, k); } /** * Recursive Problem Transformation: * * Generate all combinations of length k of the characters in rest prefixed * with the characters in…Write a Java program: A palindromic number or palindrome number is a number that remains the same when its digits are reversed. Like 16461. Write a Java program that implements the helper method isPalindrome() that decides whether a given int number is a palindrome or not. Do not use ready String class functions like reverse() or arrays and array functions!!! Input FormatA line contains an Integer N. Constraints 1 <= N <= 999999 Output FormatA text containing either "This number is a palindrome" or "This number is not a palindrome" (the code should have the structure as in the picture)import java.util.*;class Arraylist{public static void main(String args[]){ArrayList obj = new ArrayList();obj.add("A");obj.add("B");obj.add("C");obj.add(1, "D");System.out.println(obj);}} What will the given code prints on the output screen.
- For any element in keysList with a value greater than 40, print the corresponding value in itemsList, followed by a semicolon (no spaces). Ex: If the input is: 32 105 101 35 10 20 30 40 the output is: 20;30; import java.util.Scanner; public class Compare { public static void main(String args[]) {final int SIZE_LIST = 4;int[] keysList = new int[SIZE_LIST];int[] itemsList = new int[SIZE_LIST];int i;Scanner input = new Scanner(System.in); keysList[0] = input.nextInt();keysList[1] = input.nextInt();keysList[2] = input.nextInt();keysList[3] = input.nextInt(); itemsList[0] = input.nextInt();itemsList[1] = input.nextInt();itemsList[2] = input.nextInt();itemsList[3] = input.nextInt(); /* Your code goes here */ System.out.println();} }public class ReplaceEvenWithZero { public static void main(String[] args) { int[]a = new int[]{14, 2, 19, 3, 15, 22, 18, 7, 44, 39, 51, 78} ; // Print array a for (int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } System.out.println(); // Replace the even elements. replaceEven(a); // Print array a again to see new elements. for (int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } System.out.println(); } /** Replace the even elements in the given array with 0 @param arr the array to use for the replacements */ public static void replaceEven(int[] arr) { //-----------Start below here. To do: approximate lines of code = 2 // Write a for loop to go through each element, determine if it is even //If so, replace with 0. Hint: use the modulus operator % to determine if an…public class Soru2 { static void sort(char a[]) { for(int i = 0; i < a.length; i++) { for(int j = i + 1; j < a.length; j++) { if(a[j] == 'X' && a[i] == 'O') { char temp = a[j]; a[j]=a[i]; a[i]=temp; } } } for(int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } } public static void main(String[] args) { char a[] = {'X','O','X','O','X','O','X','O','X','O'}; sort(a); } } Calculate the time complexity and number of steps of the algorithm?
- java: Run length coding is a method to represent a string in a more compact manner. Each character that occurs more than 2 times in a row is represented by the character and a number following it. Two examples are: "abba" → "abba""abcccbbbba" → "abc3b4a"Write a function that calculates how many characters the encoded string is shorter than the original.For any element in keysList with a value smaller than 50, print the corresponding value in itemsList, followed by a space. Ex: If the input is: 32 105 101 35 10 20 30 40 the output is: 10 40 import java.util.Scanner; public class Compare { public static void main(String args[]) { final int SIZE_LIST = 4; int[] keysList = new int[SIZE_LIST]; int[] itemsList = new int[SIZE_LIST]; int i; Scanner input = new Scanner(System.in); keysList[0] = input.nextInt(); keysList[1] = input.nextInt(); keysList[2] = input.nextInt(); keysList[3] = input.nextInt(); itemsList[0] = input.nextInt(); itemsList[1] = input.nextInt(); itemsList[2] = input.nextInt(); itemsList[3] = input.nextInt(); //answer// System.out.println(); } }Methods Details: public static java.lang.StringBuffer[] getArrayStringsLongerThan(java.lang.StringBuffer[] array, int length) Returns a StringBuffer array with COPIES of StringBuffer objects in the array parameter that have a length greater than the length parameter. If no strings are found an empty array will be returned. Parameters: array - length - Returns: Throws:java.lang.IllegalArgumentException - When a null array parameter is provided