Convert the following while-loop into an equivalent for-loop as closely as possible using the codesnippet bank and template below. It is not enough for the loop to be functionally identical; it needsto follow any patterns mentioned in lecture.int i10;while ( i >= 0 ) {i--;+ i);%3DSystem.out.println("Iteration:i2;}[ Choose ]//LEAVE BLANKfor (1.; 2.{4.continueSystem.out.println("Iterati}i -= 2%Di> 0int i = 101.i++break2.i--i>= 0i -= 3int i = 04.[ Choose ]>3.

1. int i = 10 2. i >= 0 3. i-=2 4. i--

Convert the following while-loop into an equivalent for-loop as closely as possible using the code snippet bank and template below. It is not enough for the loop to be functionally identical; it need: o follow any patterns mentioned in lecture. int i = 10; while ( i >= 0 ) { i--; System.out.println("Iteration: + i); i -= 2; } [ Choose ]

Convert the following while-loop into an equivalent for-loop as closely as possible using the code snippet bank and template below. It is not enough for the loop to be functionally identical; it need: o follow any patterns mentioned in lecture. int i = 10; while ( i >= 0 ) { i--; System.out.println("Iteration: + i); i -= 2; } [ Choose ]

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Rewrite the following piece of java code using iterator pattern. And change the operation inside the for loop to another operation of your own. And What was the advantage of using the iterator ? import java.util.Iterator String[ ] originalData = { "one", "two", "three", "four", "five" }; List<String> strings = new ArrayList<>(Arrays.asList(originalData)); for ( int i=0; i<strings.size(); i++) { // process strings.get(i): here, just printSystem.out.println(strings.get(i));}
Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print: 7 9 11 10 10 11 9 7 import java.util.Scanner; public class CourseGradePrinter {public static void main (String [] args) {Scanner scnr = new Scanner(System.in);final int NUM_VALS = 4;int [] courseGrades = new int[NUM_VALS];int i; for (i = 0; i < courseGrades.length; ++i) {courseGrades[i] = scnr.nextInt();} /* Your solution goes here */ P.S. The answer came out wrong the first try 7 9 11 10 10 11 9 7Enter
Enhance the “Random Letter Compare” program from the previous exercise, to make it a Letter Guessing Game.Modifications,- let the first random letter be the computer’s pick of random letter that the user must guessand let the guess be from A…Z (all 26 letters)- remove the display of the random letter (but see below, TESTING and DEBUGGING)- let the second letter be input from the user, so prompt the user for a guess: "Guess which letter: "o to convert console input from String to a char, consider using:guessLetter = scan.next().toUpperCase().charAt(0); // return input character- compare the user’s character against the computer’s pick, to evaluate the correct result:o if the user guessed correctly, display,“You guessed correctly!”o else, if the user’s guess is before the computer’s pick, display,“Oh, too bad, the letter is after.”o else, the user’s guess must be after the computer’s pick, so display,“Oh, too bad, the letter is before.”Run your program to test all possible…
Write Java statements that use a for each loop to cycle through all the elements in an ArrayList of doubles named grades.
*in java* Write a for loop to print all NUM_VALS elements of array hourlyTemp. Separate elements with a comma and space. Ex: If hourlyTemp = {90, 92, 94, 95}, print:90, 92, 94, 95 Your code's output should end with the last element, without a subsequent comma, space, or newline. import java.util.Scanner; public class PrintWithComma {public static void main (String [] args) {Scanner scnr = new Scanner(System.in);final int NUM_VALS = 4;int[] hourlyTemp = new int[NUM_VALS];int i; for (i = 0; i < hourlyTemp.length; ++i) {hourlyTemp[i] = scnr.nextInt();} /* Your solution goes here */ System.out.println("");}}
Answer the following questions with the best matching answer about this code public static void checkOne() { int i = 9; do { System.out.println("Line: " + i); } while (i++ < 10); System.out.println("end"); } Answers: The number of ties code in the loop is executed The second line printed Would a while loop execute given the same conditions? If we modified i++ to ++i how many times would the loop run (or no change)? Choices to match with the answers: 0 no 1 yes Line: 9 3 2 no change end
Rewrite the following loops, using the enhanced for loop construct. Here, values is an array of floating - point numbers. a. for (int i = 0; i < values.length; i++) { total = total + values[i]; } b. for (int i = 1; i < values.length; i++) { total = total + values[i]; } c. for (int i = 0; i < values.length; i++) { if (values[i] == target) { save = i; }.
._4Give back an you please give me a code that runs properly this time and give a proper output not an empty one Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.A region is captured by flipping all 'O's into 'X's in that surrounded region.For example, for the following board,X X X XX O O XX X O XX O X XAfter capturing all regions, the board becomesX X X XX X X XX X X XX O X XThe algorithm should take an input file, named “input.txt”, and output the resulting board in the console.Sample input.txt6X X X X X OX O X O O XX X O O O XX X X O X XX O X X O OX X X X X XThe first line is a number indicating the size of the board; in the above example, 6 means the board is .Output for in console:X X X X X OX X X X X XX X X X X XX X X X X XX X X X O OX X X X X X.
Suppose that we are given a divisor and a target count, and we need to find the range of numbers from 1 that will have the target count of numbers divisible by the divisor. For example. If the divisor is 3 and the target count is 4, the range is 1 to 12, as in this range there are 4 numbers divisible by 3, viz. 3, 6, 9, and 12. Use a while loop to find the ending number of the range within which we will have the target count of numbers that are divisible by the divisor and print the result. Test the code with divisor = 13 and target count = 5. The printed result should be “Between 1 and 65 there are 5 numbers divisible by 13.” In rstudio
Can you show me how you would implement Step 3 within the code please? Step 3: a=int(input()) b=int(input()) starting_point=[] starting_point.append(a) starting_point.append(b) remaining_points=[] for i in range(n): temp=[] a=int(input()) b=int(input()) temp.append(a) temp.append(b) remaining_points.append(temp) #appending each point as a list
def loopy_madness_with_while_loops(string1: str, string2: str) -> str:"""Given two strings <string1> and <string2>, return a new string thatcontains letters from these two strings "interwoven" together, starting withthe first character of <string1>. If the two strings are not of equallength, then start looping "backwards-and-forwards" in the shorter stringuntil you come to the end of the longer string."interwoven" (or "interweaving") means constructing a new string by takingthe first letter from the first string, adding the first letter of thesecond string, adding the second letter of the first string,adding the second letter of the second string, and so on."backwards-and-forwards" is a custom looping term. First the loop startsat position 1 (index 0) and goes until position n (i.e., the end). Once theloop reaches position n, it goes backwards, starting at position n - 1 andgoes to position 1 (index 0). This repeats until the two strings areinterwoven. For example,…
reversing the elements of an array involves swapping the corespodning elements of the array: the first with the last, the second with the next to the last, and so on, all the way to the middle of the array. given an array a, an int variable n containing the number of elements in a, and two other int variables, k and temp, write a loop that reverses the elements of the array. do not use any other variables besides a, n, k, and temp. in c++