Unit 4 - Iteration:
- This is the homework quiz for unit 4, iterations
- 4 multiple choice questions
- 2 programming hacks
- 1 bonus programming hack (required to get above 0.9)
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
Click to reveal answer:
DExplain your answer. (explanation is graded not answer)
for (int i = 3; i <= 12; i++) {
System.out.print(i + " ");
}
The answer is D because the for loop is iterating from the numbers 3 to 12 inclusive and “i” is being post-incremented by 1 each time. Therefore it prints out 3 4 5 6 7 8 9 10 11 12
Bonus:
- Explain the difference between using a variable like i inside a for loop, vs. using a variable that exists in the code itself for a while loop
Question 2:
How many times does the following method print a “*” ?
A. 9
B. 7
C. 8
D. 6
Click to reveal answer:
CExplain your answer. (explanation is graded not answer)
for (int i = 3; i < 11; i++) {
System.out.print("*");
}
C. It prints out 8 stars because i starts from 3 and ends at 10.
10-3+1 = 8.
Question 3:
What does the following code print?
A. -4 -3 -2 -1 0
B. -5 -4 -3 -2 -1
C. 5 4 3 2 1
Click to reveal answer:
AExplain your answer. (explanation is graded not answer)
int x = -5;
while (x < 0)
{
x++;
System.out.print(x + " ");
}
A. x starts at -5 and ends at -1. Since x is incremented by 1 each iteration, it will print out all the numbers from -5 to -1 inclusive.
So, the output is: -5 -4 -3 -2 -1
Question 4:
What does the following code print?
A. 20
B. 21
C. 25
D. 30
Click to reveal answer:
BExplain your answer. (explanation is graded not answer)
int sum = 0;
for (int i = 1; i <= 5; i++) {
if (i % 2 == 0) {
sum += i * 2;
} else {
sum += i;
}
}
System.out.println(sum);
When i is even, i is doubled and added to the sum.
When i is odd, i is just added to the sum normally.
When i = 1: it’s odd, so sum += 1 → sum = 1.
When i = 2: it’s even, so sum += 2 * 2 → sum = 1 + 4 = 5.
When i = 3: it’s odd, so sum += 3 → sum = 5 + 3 = 8.
When i = 4: it’s even, so sum += 4 * 2 → sum = 8 + 8 = 16.
When i = 5: it’s odd, so sum += 5 → sum = 16 + 5 = 21.
Therefore, the answer is B. 21
Loops HW 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
// Using while loop:
import java.util.ArrayList;
public class myWhileLoop {
public static void main(String[] args) {
ArrayList<Integer> divisibleBy3Or5 = new ArrayList<>();
int i = 1;
while (i <= 50) {
if (i % 3 == 0 || i % 5 == 0) {
divisibleBy3Or5.add(i);
}
i++;
}
System.out.println("Numbers divisible by 3 or 5 (while loop): " + divisibleBy3Or5);
}
}
myWhileLoop.main(null);
Numbers divisible by 3 or 5 (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]
// Using a for loop
import java.util.ArrayList;
public class myForLoop {
public static void main(String[] args) {
ArrayList<Integer> divisibleBy3Or5 = new ArrayList<>();
for (int i = 1; i <= 50; i++) {
if (i % 3 == 0 || i % 5 == 0) {
divisibleBy3Or5.add(i);
}
}
System.out.println("Numbers divisible by 3 or 5 (for loop): " + divisibleBy3Or5);
}
}
myForLoop.main(null);
Numbers divisible by 3 or 5 (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;
import java.util.List;
public class PalindromeFinder {
public static void main(String[] args) {
// Sample Input
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};
// List to store palindromes
List<Integer> palindromes = new ArrayList<>();
int index = 0;
while (index < testList.length) {
if (isPalindrome(testList[index])) {
palindromes.add(testList[index]);
}
index++;
}
// Output the result
System.out.println(palindromes);
}
// Function to check if a number is a palindrome
public static boolean isPalindrome(int num) {
String strNum = String.valueOf(num);
String reversedStrNum = new StringBuilder(strNum).reverse().toString();
return strNum.equals(reversedStrNum);
}
}
Bonus Hack (for above 0.9)
Use a for loop to output a spiral matrix with size n
Example:
Sample Input: n = 3
Output: [[1, 2, 3], [8, 9, 4], [7, 6, 5]]
public class SpiralMatrix {
public static void main(String[] args) {
// Sample Input
int n = 3;
// Generate the spiral matrix
int[][] spiralMatrix = generateSpiralMatrix(n);
// Output the result
printMatrix(spiralMatrix);
}
public static int[][] generateSpiralMatrix(int n) {
int[][] matrix = new int[n][n]; // Initialize the n x n matrix
int top = 0, bottom = n - 1, left = 0, right = n - 1; // Boundaries
int num = 1; // Starting number
while (top <= bottom && left <= right) {
// Traverse from left to right
for (int i = left; i <= right; i++) {
matrix[top][i] = num++;
}
top++; // Move the top boundary down
// Traverse from top to bottom
for (int i = top; i <= bottom; i++) {
matrix[i][right] = num++;
}
right--; // Move the right boundary left
// Traverse from right to left (if applicable)
if (top <= bottom) {
for (int i = right; i >= left; i--) {
matrix[bottom][i] = num++;
}
bottom--; // Move the bottom boundary up
}
// Traverse from bottom to top (if applicable)
if (left <= right) {
for (int i = bottom; i >= top; i--) {
matrix[i][left] = num++;
}
left++; // Move the left boundary right
}
}
return matrix;
}
public static void printMatrix(int[][] matrix) {
for (int[] row : matrix) {
System.out.print("[");
for (int j = 0; j < row.length; j++) {
System.out.print(row[j]);
if (j < row.length - 1) {
System.out.print(", ");
}
}
System.out.println("]");
}
}
}
SpiralMatrix.main(null);
[1, 2, 3]
[8, 9, 4]
[7, 6, 5]