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: D

Explain your answer. (explanation is graded not answer)

for (int i = 3; i <= 12; i++) {
   System.out.print(i + " ");
}

It would print D becuase it starts with 3 and it’s “i=3” and then it would keep adding 1 until it gets to 12 becuase it says <= to 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: C

Explain your answer. (explanation is graded not answer)

for (int i = 3; i < 11; i++) {
   System.out.print("*");
}

It’s C because it prints out 8 times because if it starts with 3 it would print an astrik until it hits 10.

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: A

Explain your answer. (explanation is graded not answer)

int x = -5;
while (x < 0)
{
   x++;
   System.out.print(x + " ");
}

A because it continuously adds 1 to it; nevertheless, it stops at 0 because x is no longer less than 0.

Question 4:

What does the following code print?

A. 20

B. 21

C. 25

D. 30

Click to reveal answer: B

Explain 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);

B because 21 will be printed by the code. For each value of i, the addition process starts at 1 and continues with 4 (double of 2) for i = 2, 3 for i = 3, 8 (double of 4) for i = 4, and lastly 5 for i = 5. There is a total of 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
public class Main {
    public static void main(String[] args) {
        ArrayList<Integer> divisibleBy3Or5While = new ArrayList<>();
        int i = 1;
        
        while (i <= 50) {
            if (i % 3 == 0 || i % 5 == 0) {
                divisibleBy3Or5While.add(i);
            }
            i++;
        }
        
        System.out.println("Numbers divisible by 3 or 5 (while loop): " + divisibleBy3Or5While);
    }
}
Main.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]

public class Main {
    public static void main(String[] args) {
        ArrayList<Integer> divisibleBy3Or5For = new ArrayList<>();
        
        for (int i = 1; i <= 50; i++) {
            if (i % 3 == 0 || i % 5 == 0) {
                divisibleBy3Or5For.add(i);
            }
        }
        
        System.out.println("Numbers divisible by 3 or 5 (for loop): " + divisibleBy3Or5For);
    }
}
Main.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;

public class Main {
    public static void main(String[] args) {
        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 i = 0;
        while (i < testList.length) {
            if (isPalindrome(testList[i])) {
                palindromes.add(testList[i]);
            }
            i++;
        }
        System.out.println("Palindromes: " + palindromes);
    }
    public static boolean isPalindrome(int number) {
        int original = number;
        int reversed = 0;
        
        while (number != 0) {
            int digit = number % 10;
            reversed = reversed * 10 + digit;
            number /= 10;
        }
        
        return original == reversed;
    }
}
Main.main(null)

Palindromes: [4444, 515, 2882, 6556, 595]

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) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter matrix size n: ");
        int n = scanner.nextInt();

        int[][] matrix = generateSpiralMatrix(n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(matrix[i][j] + " ");
            }
            System.out.println();
        }
        scanner.close();
    }

    public static int[][] generateSpiralMatrix(int n) {
        int[][] matrix = new int[n][n];
        int value = 1;
        int top = 0, bottom = n - 1;
        int left = 0, right = n - 1;

        while (top <= bottom && left <= right) {
            for (int i = left; i <= right; i++) {
                matrix[top][i] = value++;
            }
            top++;
            for (int i = top; i <= bottom; i++) {
                matrix[i][right] = value++;
            }
            right--;
            if (top <= bottom) {
                for (int i = right; i >= left; i--) {
                    matrix[bottom][i] = value++;
                }
                bottom--;
            }
            if (left <= right) {
                for (int i = bottom; i >= top; i--) {
                    matrix[i][left] = value++;
                }
                left++;
            }
        }

        return matrix;
    }
}
SpiralMatrix.main(null)

Enter matrix size n: 1 2 3 
8 9 4 
7 6 5