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 + " ");
}

D

the for loop starts with i = 3 and runs until i <= 12, so itll iterate from 3-12 and print the value of i each time and will add a space each time. sothe output is : 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

in a for loop, the variable (i) is mostly defined and controlled by the loop itself, with initialization, condition, and increment. this limits the variable to the loop. however, in a while loop, the control variable usually is outside the loop, and its initialization, condition checking, and updating are handled separately

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("*");
}

C

The for loop starts at i = 3 and runs as long as i < 11, so itll go till i = 3,4,5,6,7,8,9,10. this is 8 iterations so itll print 8 asterisks

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

the variable x is initialized to -5. the while loop continues as long as x < 0. in the loop, x++ increments x by 1, then the value of x is printed. The values printed are : -4 -3 -2 -1 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);

the loop runs from i = 1 to i = 5. The if condition checks whether i is even (i % 2 == 0). if it is, the loop adds i * 2 to sum or it just adds to i

i = 1: odd, so sum += 1 → sum = 1 i = 2: even, so sum += 2 * 2 → sum = 1 + 4 = 5 i = 3: odd, so sum += 3 → sum = 5 + 3 = 8 i = 4: even, so sum += 4 * 2 → sum = 8 + 8 = 16 i = 5: odd, so sum += 5 → sum = 16 + 5 = 21

final value is 21 lol

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

        ArrayList<Integer> divisibleNumbersWhile = new ArrayList<>();
        int i = 1;
        
        while (i <= 50) {
            if (i % 3 == 0 || i % 5 == 0) {
                divisibleNumbersWhile.add(i);
            }
            i++;
        }
        
        System.out.println("this is using while loop " + divisibleNumbersWhile);


this is 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]


        ArrayList<Integer> divisibleNumbersFor = new ArrayList<>();
        
        for (int i = 1; i <= 50; i++) {
            if (i % 3 == 0 || i % 5 == 0) {
                divisibleNumbersFor.add(i);
            }
        }
        
        System.out.println("this is using for loop" + divisibleNumbersFor);


this is 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 Main {
    public static void main(String[] args) {
        int[] 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};
        ArrayList<Integer> palindromes = new ArrayList<>();
        
        int i = 0;
        
        while (i < test_list.length) {
            if (isPalindrome(test_list[i])) {
                palindromes.add(test_list[i]);
            }
            i++;
        }
        
        System.out.println("palindromes: " + palindromes);
    }
    
    public static boolean isPalindrome(int number) {
        int original = number;
        int reverse = 0;
        
        while (number > 0) {
            int digit = number % 10;
            reverse = reverse * 10 + digit;
            number /= 10;
        }
        
        return original == reverse;
    }
}

outputs [4444, 515, 2882, 6556, 595]

(code runs and prints check, so it works, doesnt output cuz jupyter terminal being weird with the public class and public static etc… )

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]]