2.9 Implementing Selection and Iteration Algorithms

1. Example 1

   Write a method: counting multiples

   Write a method countMultiples(int low, int high, int d) that returns how many integers from low to high, inclusive, are divisible by d. Assume d is positive. For example, countMultiples(10, 30, 7) returns 3, for 14, 21 and 28. A full solution:Javapublic static int countMultiples(int low, int high, int d) { int count = 0; for (int n = low; n <= high; n++) { if (n % d == 0) { count++; } } return count; }

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   1. 1Step 1: This is the counting pattern: a counter that starts at 0, a loop over every value, and an if that adds 1 when the condition holds.
   2. 2Step 2: "Inclusive" means high must be checked too, so the condition is n <= high, not n < high.
   3. 3Step 3: The condition is divisibility: n % d == 0.
   4. 4Step 4: Return the count after the loop ends, not inside it.
   5. 5Step 5: Check: from 10 to 30, the multiples of 7 are 14, 21 and 28, so the method returns 3.

   Answer: The method above; countMultiples(10, 30, 7) returns 3.

2. Example 2

   Write a method: largest digit

   Write a method largestDigit(int num) that returns the largest digit in a non-negative integer. For example, largestDigit(4829) returns 9. One solution:Javapublic static int largestDigit(int num) { int largest = num % 10; while (num > 0) { int digit = num % 10; if (digit > largest) { largest = digit; } num = num / 10; } return largest; }

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   1. 1Step 1: Combine the digit pattern with the maximum pattern.
   2. 2Step 2: Start largest at a real digit, the last one (num % 10), so it's never an invented value.
   3. 3Step 3: Each pass takes the last digit with % 10, compares it with largest, then drops it with / 10.
   4. 4Step 4: Trace 4829: the digits come out as 9, 2, 8, 4. largest starts at 9 and nothing beats it.
   5. 5Step 5: Edge cases: largestDigit(7) is 7 and largestDigit(50) is 5. For 0, the loop never runs, and the method correctly returns 0 % 10, which is 0.

   Answer: The method above; it returns 9 for 4829, 7 for 7 and 5 for 50.

3. Example 3

   The average trap

   What does this code print?Javaint sum = 0; int count = 0; for (int n = 1; n <= 10; n++) { if (n % 3 == 0) { sum += n; count++; } } double average = (double) sum / count; System.out.println(sum + " " + count + " " + average);

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   1. 1Step 1: The loop checks 1 through 10 and keeps the multiples of 3: 3, 6 and 9.
   2. 2Step 2: sum is 18 and count is 3.
   3. 3Step 3: (double) sum / count casts sum to 18.0 first, so the division is 18.0 / 3, which is 6.0. Without the cast, sum / count would be the int 6 and print as 6.

   Answer: It prints 18 3 6.0.

Common mistakes

• !Starting a maximum at 0. If all values are negative, the answer is wrong. Start with a real value from the data.
• !Dividing two ints for an average and losing the decimal part. Cast one of them to double first.
• !Returning from inside the loop too early, before every value has been checked.
• !Writing % when you meant /, or the other way around, when pulling out digits.

On the exam

• ✓Since the 2025 course update, free-response Question 1 (Methods and Control Structures) is worth 7 points instead of 9: 4 for Part A and 3 for Part B. Part A usually needs one of these loop patterns. Name the pattern first, then write it.
• ✓The exam is now digital, so you type your free-response answers in Bluebook instead of writing them by hand. Nothing runs your code for you, so trace it once with the example from the question before you move on.