- Fun While Loop Hack:
- Fun For Loop Hack:
- Popcorn Hack:
- Homework Hack!
- Fun Stubstring Hack:
- What is wrong with this code cell(Hack)
- Loops homework hack
- Easy Hack
- Harder Hack
- Coding Hack(For above 0.9):
Fun While Loop Hack:
- find and fix the missing increment in the while loop
int i = 0;
while (i < 5) {
System.out.println(i);
i++; // Incrementing i to avoid an infinite loop
}
0
1
2
3
4
Fun For Loop Hack:
Create a program that iterates through a list of numbers (int_list = {0, 4, 51, 83, 92, 10, 123, 145}) using both a for loop and a for each loop, then split the numbers in the list into even/odd lists, and output them.
import java.util.ArrayList;
public class FunForLoopHack {
public static void main(String[] args) {
// Initial list of numbers
int[] int_list = {0, 4, 51, 83, 92, 10, 123, 145};
// Creating separate lists for even and odd numbers
ArrayList<Integer> evenNumbers = new ArrayList<>();
ArrayList<Integer> oddNumbers = new ArrayList<>();
// Using a regular for loop to iterate and split numbers
System.out.println("Using for loop:");
for (int i = 0; i < int_list.length; i++) {
int num = int_list[i];
System.out.println(num);
if (num % 2 == 0) {
evenNumbers.add(num);
} else {
oddNumbers.add(num);
}
}
// Clear the lists for the next approach (optional)
evenNumbers.clear();
oddNumbers.clear();
// Using a for-each loop to iterate and split numbers
System.out.println("\nUsing for-each loop:");
for (int num : int_list) {
System.out.println(num);
if (num % 2 == 0) {
evenNumbers.add(num);
} else {
oddNumbers.add(num);
}
}
// Output the even and odd lists
System.out.println("\nEven numbers: " + evenNumbers);
System.out.println("Odd numbers: " + oddNumbers);
}
}
FunForLoopHack.main(null);
Using for loop:
0
4
51
83
92
10
123
145
Using for-each loop:
0
4
51
83
92
10
123
145
Even numbers: [0, 4, 92, 10]
Odd numbers: [51, 83, 123, 145]
Popcorn Hack:
Iterate through the characters a string with a while loop
public class PopcornHack {
public static void main(String[] args) {
String text = "Popcorn Hack";
int index = 0;
// Using a while loop to iterate through the string
while (index < text.length()) {
// Print each character
System.out.println(text.charAt(index));
index++; // Move to the next character
}
}
}
PopcornHack.main(null);
P
o
p
c
o
r
n
H
a
c
k
Homework Hack!
code a caesar cipher that will encrypt any string with any key provided. your code should go into the encrypt() method, and should successfully pass the test cases provided as a bonus, try to use StringBuilder
public class CaesarCipher {
private int key;
private String phrase;
public CaesarCipher(int key, String phrase) {
this.key = key;
this.phrase = phrase;
}
public String encrypt() {
StringBuilder encrypted = new StringBuilder();
for (int i = 0; i < phrase.length(); i++) {
char currentChar = phrase.charAt(i);
if (Character.isLetter(currentChar)) {
char base = Character.isLowerCase(currentChar) ? 'a' : 'A';
// Shift the character within the alphabet range
char shiftedChar = (char) ((currentChar - base + key) % 26 + base);
encrypted.append(shiftedChar);
} else {
// If it's not a letter, keep it as is
encrypted.append(currentChar);
}
}
return encrypted.toString();
}
public static void main(String[] args) {
CaesarCipher test1 = new CaesarCipher(3, "hello world");
CaesarCipher test2 = new CaesarCipher(10, "abcdefg");
CaesarCipher test3 = new CaesarCipher(20, "i love csa");
System.out.println("test 1: " + test1.encrypt()); // khoor zruog
System.out.println("test 2: " + test2.encrypt()); // klmnopqrst
System.out.println("test 3: " + test3.encrypt()); // c fihy wmu
}
}
CaesarCipher.main(null);
test 1: khoor zruog
test 2: klmnopq
test 3: c fipy wmu
Fun Stubstring Hack:
Create a program that scrables any word that you put in, by reversing the first and last letters, then reversing the center letters. (example: sold becomes dlos, computer becomes retupmoc)
import java.util.Scanner;
public class FunStubstringHack {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
// Prompt user for input
System.out.println("Enter a word to scramble: ");
String word = scanner.nextLine();
// Scramble the word
String scrambledWord = scrambleWord(word);
// Output the result
System.out.println("Scrambled word: " + scrambledWord);
}
public static String scrambleWord(String word) {
// Handle words with less than 2 characters
if (word.length() < 2) {
return word; // No scrambling needed
}
// Get the first and last letters
char firstLetter = word.charAt(0);
char lastLetter = word.charAt(word.length() - 1);
// Get the center letters
String centerLetters = word.substring(1, word.length() - 1);
// Reverse the center letters
String reversedCenter = new StringBuilder(centerLetters).reverse().toString();
// Construct the scrambled word
return lastLetter + reversedCenter + firstLetter;
}
}
FunStubstringHack.main(null);
Enter a word to scramble:
Scrambled word: rebotco
What is wrong with this code cell(Hack)
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
System.out.print("Enter the number of rows: ");
int rows = scanner.nextInt();
for (int i = rows; i>1; i-) {
// Print numbers from 1 to i
for (int j = 1; j <= i; j++) {
System.out.print(j + " ");
}
// Move to next line after each row
System.out.println();
}
scanner.close();
Answer: In the loop for (int i = rows; i > 1; i-), the decrement operator is incorrectly written as i-. It should be i-- to properly decrement the value of i in each iteration. This causes the loop not to execute as expected.
Answer: B. 20
Explanation: The loop will run 5 times because 8-3 is 5. Then in each loop, there are more loops, each time 4 loops because 5-1 is 4. If i do 5 times 4, I get 20, which is the answer.
This will print * 24 times in a new line because the outer loop runs 6 times and the inner loop runs 4 times, so the code will be run 24 times, giving us * 24 times.
Question 1:
What does the following code print?
A. 5 6 7 8 9
B. 4 5 6 7 8 9 10 11 12
C. 3 5 7 9 11
D. 3 4 5 6 7 8 9 10 11 12
for (int i = 3; i <= 12; i++)
{
System.out.print(i + " ");
}
Answer: D
Question 2:
How many times does the following method print a *?
A. 9
B. 7
C. 6
D. 10
for (int i = 3; i < 9; i++)
{
System.out.print("*");
}
Answer: C
Question 3:
What does the following code print?
A. 5 4 3 2 1
B. -5 -4 -3 -2 -1
C. -4 -3 -2 -1 0
int x = -5;
while (x < 0)
{
x++;
System.out.print(x + " ");
}
Answer: C
Loops homework hack
Easy Hack
- use a while loop to find the numbers from 1-50 that are divisible by 3 or 5, then store them into a list (make sure to print it out at the end)
- use a for loop to do the same thing detailed above
import java.util.ArrayList;
public class DivisibleByThreeOrFiveWhile {
public static void main(String[] args) {
ArrayList<Integer> divisibleNumbers = new ArrayList<>();
int i = 1; // Start from 1
// Use a while loop to find numbers from 1 to 50
while (i <= 50) {
if (i % 3 == 0 || i % 5 == 0) {
divisibleNumbers.add(i);
}
i++; // Increment i
}
// Print the result
System.out.println("Numbers from 1 to 50 that are divisible by 3 or 5 (using while loop): " + divisibleNumbers);
}
}
DivisibleByThreeOrFiveWhile.main(null);
Numbers from 1 to 50 that are divisible by 3 or 5 (using while loop): [3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50]
import java.util.ArrayList;
public class DivisibleByThreeOrFiveFor {
public static void main(String[] args) {
ArrayList<Integer> divisibleNumbers = new ArrayList<>();
// Use a for loop to find numbers from 1 to 50
for (int i = 1; i <= 50; i++) {
if (i % 3 == 0 || i % 5 == 0) {
divisibleNumbers.add(i);
}
}
// Print the result
System.out.println("Numbers from 1 to 50 that are divisible by 3 or 5 (using for loop): " + divisibleNumbers);
}
}
DivisibleByThreeOrFiveFor.main(null);
Numbers from 1 to 50 that are divisible by 3 or 5 (using for loop): [3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50]
Harder Hack
Palindromes are numbers that have the same value when reversed (ex: “123321” or “323”). Create a program that uses a while loop that outputs all palindromes in any given list.
Sample Input: test_list = [5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770, 4442, 913, 2508, 1116, 9969, 9091, 522, 8756, 9527, 7968, 1520, 4444, 515, 2882, 6556, 595]
Sample Output: 4444, 515, 2882, 6556, 595
import java.util.ArrayList;
public class PalindromeFinder {
public static void main(String[] args) {
// Sample input list
int[] testList = {5672, 235, 5537, 6032, 317, 8460, 1672, 8104, 7770,
4442, 913, 2508, 1116, 9969, 9091, 522, 8756,
9527, 7968, 1520, 4444, 515, 2882, 6556, 595};
ArrayList<Integer> palindromes = new ArrayList<>();
int index = 0;
// Use a while loop to iterate through the testList
while (index < testList.length) {
if (isPalindrome(testList[index])) {
palindromes.add(testList[index]);
}
index++; // Increment index
}
// Print the result
System.out.println("Palindromes in the list: " + palindromes);
}
// Method to check if a number is a palindrome
public static boolean isPalindrome(int number) {
String strNumber = Integer.toString(number); // Convert number to string
int left = 0;
int right = strNumber.length() - 1;
// Check if the string is a palindrome
while (left < right) {
if (strNumber.charAt(left) != strNumber.charAt(right)) {
return false; // Not a palindrome
}
left++;
right--;
}
return true; // It's a palindrome
}
}
PalindromeFinder.main(null);
Palindromes in the list: [4444, 515, 2882, 6556, 595]
Coding Hack(For above 0.9):
Use a for loop to output a spiral matrix with size n
For example:
Sample Input: 3
Output: [[1, 2, 3], [8, 9, 4], [7, 6, 5]]
import java.util.Arrays;
public class SpiralMatrix {
public static void main(String[] args) {
int n = 3; // Sample input size
int[][] matrix = new int[n][n]; // Create a 2D array for the spiral matrix
int top = 0, bottom = n - 1; // Initialize top and bottom boundaries
int left = 0, right = n - 1; // Initialize left and right boundaries
int num = 1; // Start filling the matrix from 1
// Fill the matrix in a spiral order
for (int i = 0; i < n * n; i++) {
if (top <= bottom) {
for (int j = left; j <= right; j++) {
matrix[top][j] = num++;
}
top++; // Move the top boundary down
}
if (left <= right) {
for (int j = top; j <= bottom; j++) {
matrix[j][right] = num++;
}
right--; // Move the right boundary left
}
if (top <= bottom) {
for (int j = right; j >= left; j--) {
matrix[bottom][j] = num++;
}
bottom--; // Move the bottom boundary up
}
if (left <= right) {
for (int j = bottom; j >= top; j--) {
matrix[j][left] = num++;
}
left++; // Move the left boundary right
}
}
// Print the spiral matrix
System.out.println("Spiral Matrix of size " + n + ":");
for (int[] row : matrix) {
System.out.println(Arrays.toString(row));
}
}
}
SpiralMatrix.main(null);
Spiral Matrix of size 3:
[1, 2, 3]
[8, 9, 4]
[7, 6, 5]