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AP CSA Unit 4.3 Practice: Loop Algorithms MCQs & Exam Prep

Home » AP CSA MCQ » Unit 4 » Loop Algorithms

AP CSA Practice: Loop Algorithms & String Processing

📚 Topic 4.3 ⏱️ 10 Min Practice ✅ Updated for 2027 Exam

Once you master basic loop syntax, the AP Computer Science A exam tests how you apply loops to solve standard algorithmic problems. Topic 4.3 focuses on common computational patterns including accumulator variables, string manipulation, character traversal, and digit extraction.

In this interactive practice set, we cover Topic 4.3: Developing Algorithms Using Loops. These 10 multiple-choice questions challenge your ability to trace state changes in strings and numbers during repetitive execution.

Key Concepts to Remember Before You Start:

  • Accumulator Patterns: Initializing sums to 0 or products to 1 before a loop, then updating them inside the loop body.
  • String Processing: Traversing strings using substring(i, i + 1) combined with loop bounds like s.length().
  • Reversing Strings: Building new strings backward by looping from s.length() - 1 down to 0.
  • Digit Extraction: Isolating the rightmost digit of a number using modulo (% 10) and stripping it using integer division (/ 10).

Test your knowledge with the interactive quiz below. Review the explanations for any questions you miss to sharpen your algorithm tracing skills!

AP CSA 4.3: Loop Algorithms

Question 1 of 10 | Score: 0
1. Consider the following code segment designed to process a string:
String str = “java”;
String rev = “”;
for (int i = str.length() – 1; i >= 0; i–) {
    rev += str.substring(i, i + 1);
}
System.out.println(rev);
Correct Answer: B Trace breakdown: str.length() is 4, so i starts at index 3 (‘a’). The loop steps backward to 2 (‘v’), 1 (‘a’), and 0 (‘j’). Concatenating these characters builds the reversed string “avaj”.
2. Consider the following code segment:
String s = “code.org”;
int count = 0;
for (int i = 0; i < s.length(); i++) {
    if (s.substring(i, i + 1).equals(“o”)) {
        count++;
    }
}
System.out.println(count);
Correct Answer: B The loop checks every character in “code.org”. The letter ‘o’ appears twice—once in “code” and once in “.org”. Thus, the accumulator count increments to 2.
3. Consider the following code segment that processes digits:
int n = 4825;
int maxDigit = 0;
while (n > 0) {
    int d = n % 10;
    if (d > maxDigit) {
        maxDigit = d;
    }
    n /= 10;
}
System.out.println(maxDigit);
Correct Answer: C This algorithm finds the maximum digit in n. Using modulo (% 10) and integer division (/ 10), it extracts digits from right to left: 5, then 2, then 8, and finally 4. The highest value encountered and stored in maxDigit is 8.
4. Consider the following code segment:
String word = “computer”;
String result = “”;
for (int i = 1; i < word.length(); i += 2) {
    result += word.substring(i, i + 1);
}
System.out.println(result);
Correct Answer: B The loop starts at index 1 and increments by 2, extracting characters at odd indices:
– Index 1: ‘o’
– Index 3: ‘p’
– Index 5: ‘t’
– Index 7: ‘r’
Combining these gives the string “optr”.
5. Consider the following code segment using an accumulator:
int sum = 0;
for (int i = 1; i <= 5; i++) {
    sum += (i * 10);
}
System.out.println(sum);
Correct Answer: C The loop runs 5 times. During each iteration, it adds i * 10 to sum.
Iteration values added: 10 + 20 + 30 + 40 + 50 = 150.
6. Consider the following code segment:
int n = 1234;
int sum = 0;
while (n > 0) {
    sum += n % 10;
    n /= 10;
}
System.out.println(sum);
Correct Answer: A This is the standard algorithm for finding the sum of all digits in an integer. It extracts the digits 4, 3, 2, and 1, adding them sequentially to the sum accumulator. 4 + 3 + 2 + 1 = 10.
7. Consider the following code segment:
String s = “apple”;
String res = “”;
for (int i = 0; i < s.length(); i++) {
    String c = s.substring(i, i + 1);
    if (res.indexOf(c) < 0) {
        res += c;
    }
}
System.out.println(res);
Correct Answer: C This algorithm removes duplicate characters while preserving original order. indexOf(c) returns -1 (which is < 0) if a character is not yet in res. It appends ‘a’, ‘p’, ‘l’, and ‘e’. When it encounters the second ‘p’, indexOf("p") returns 1, so it is ignored. Output: “aple”.
8. Consider the following code segment:
int val = 0;
for (int i = 1; i <= 4; i++) {
    val = val * 2 + i;
}
System.out.println(val);
Correct Answer: B Trace table:
– Start: val = 0
– i = 1: val = (0 * 2) + 1 = 1
– i = 2: val = (1 * 2) + 2 = 4
– i = 3: val = (4 * 2) + 3 = 11
– i = 4: val = (11 * 2) + 4 = 26
The loop terminates, printing 26.
9. Consider the following code segment:
String s = “abracadabra”;
int c = 0;
for (int i = 0; i < s.length() – 1; i++) {
    if (s.substring(i, i + 2).equals(“ab”)) {
        c++;
    }
}
System.out.println(c);
Correct Answer: B The loop bounds (i < s.length() - 1) safely allow substring(i, i + 2) to grab pairs of characters without crashing. It scans for the exact string “ab”, which appears twice in “abracadabra” (at the very beginning and near the end).
10. Which of the following mathematical expressions is universally used within a loop to isolate the rightmost digit of a positive integer n?
Correct Answer: B The modulo operator (%) returns the remainder of a division. Dividing any base-10 number by 10 leaves the rightmost digit as the remainder (e.g., 345 % 10 = 5). Using n / 10 chops off the rightmost digit instead of isolating it.

Unit 4.3 Algorithms Practice Complete!

0 / 10

Why String Algorithms Dominate the FRQ Section

While you will see loop algorithms throughout the multiple-choice section, they are absolutely guaranteed to appear on the Free Response Questions (FRQs). The College Board frequently requires you to iterate through an entire string to find specific patterns, count occurrences of a target word, or build a newly modified string character-by-character.

Ready for the Next Topic?

Now that you know how to process single sequences and simple algorithms, it is time to look at what happens when loops are placed inside other loops.

Continue your AP CSA review by proceeding to Topic 4.4: Nested Iteration Practice, where we explore inner-to-outer execution limits and matrix logic.

← Previous: Topic 4.2 For Loops
Next: Topic 4.4 Nested Loops →

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