Unit 4 - HW Quiz | Nighthawk Pages

It’s D because we can first eliminate A and B because those both didn’t start with a 3. Then we can eliminate C because in D, the code prints the numbers that is starting from 3 and incrementing by 1 and then end with a 12, unlike in C, it skips part of the numbers from 3 to 12.

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

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:

Explain your answer. (explanation is graded not answer)

⬆️⬆️⬆️ Answering

It’s C beacuse the loop runs from i=3 to i=10, which means that there will be total of 8 *s. Then each iteration corresponds to one increment, there fore, the loop iterates a total of 8 times.

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

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:

Explain your answer. (explanation is graded not answer)

⬆️⬆️⬆️ Answering

The correct answer will be A since after the first iteration, x is -5 then after the incrementing, the x will becomes -4. The only answer choice that has -4 starting is only option A. Therefore the correct answer is A.

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

Question 4:

What does the following code print?

A. 20

B. 21

C. 25

D. 30

Click to reveal answer:

Explain your answer. (explanation is graded not answer)

⬆️⬆️⬆️ Answering

The correct answer is B. This is because that the loop check each value of i from 1 –> 5, and then adding either i itself for the odd numbers or double the value of the i for even numbers. Then it will be 21 in total after a calculation, so therefore, the correct option is B.

int sum = 0;

for (int i = 1; i <= 5; i++) {
    if (i % 2 == 0) {
        sum += i * 2;
    } else {
        sum += i;
    }
}

System.out.println(sum);

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

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

usingForLOop.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]

⬆️⬆️⬆️ Answering

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

public class PalindromeFinder {
    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};
        
        // creating a new arraylist to store the palindromes
        ArrayList<Integer> palindromes = new ArrayList<>();
        
        int i = 0;
        
        while (i < testList.length) {
            if (isPalindrome(testList[i])) {
                palindromes.add(testList[i]); 
            }
            i++; // The increment the index
        }
        
        // printing the palindromes
        for (int num : palindromes) {
            System.out.print(num + " ");
        }
    }
    
    // the helper method to check if a number is a palindrome or not
    public static boolean isPalindrome(int num) {
        int originalNum = num;
        int reversedNum = 0;
        
        // Reverse the number
        while (num > 0) {
            int digit = num % 10;
            reversedNum = reversedNum * 10 + digit;
            num /= 10;
        }
        
        // checking if the original number is = to the reversed number
        return originalNum == reversedNum;
    }
}


PalindromeFinder.main(null);

4444 515 2882 6556 595

⬆️⬆️⬆️ Answering

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]]