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WORKBOOK · FREE SAMPLE BELOW

Java interview practice workbook

Ten original coding tasks covering methods, arrays, object behaviour, collections and input handling. Work through each task, then compare with a complete Java 17 solution.

Try the free sampleAsk about the full workbook

The team will confirm price, access and delivery before purchase.

Who it is for

Learners who can run a Java class and want to explain their decisions in an interview.

Attempt each task before looking at its answer. State the assumptions, predict the result and test one boundary case. A working answer is the start: explain why it behaves correctly and what you would change if the requirements changed.

This is a self-study resource. It is not an official exam, a qualification or a promise of employment. If you need guided classes, ask the team about the relevant course separately.

What is included

  • 10 original coding tasks
  • Complete runnable Java 17 solutions
  • Expected output and review points
  • A self-assessment rubric and interview follow-up prompts

Assess each task against its result, edge cases, explanation and readability. Record a mistake and retry the task later without opening the solution.

Try two free sample tasks

These tasks show the format and level of the full workbook. Read the prompt first; open the answer only after making your own attempt.

1. Sum the even values

Write sumEven(int[] values). Return the sum of even values, including zero and negative even numbers. Return zero for an empty array. Demonstrate it with {3, 6, -2, 0, 5}.

Compare the answer and review points
Main.java
public class Main {
    static int sumEven(int[] values) {
        int sum = 0;
        for (int value : values) if (value % 2 == 0) sum += value;
        return sum;
    }
    public static void main(String[] args) {
        System.out.println(sumEven(new int[]{3, 6, -2, 0, 5}));
        System.out.println(sumEven(new int[]{}));
    }
}

Expected result

4
0
  • Handles negative even numbers
  • Returns zero for empty input
  • Keeps the accumulator outside the loop

2. Find the first largest-mark position

Return the index of the first largest mark, or -1 for an empty array. For {70, 95, 95, 82}, the answer must be 1 rather than 2. Do not sort or mutate the array.

Compare the answer and review points
Main.java
public class Main {
    static int firstLargest(int[] marks) {
        if (marks.length == 0) return -1;
        int best = 0;
        for (int i = 1; i < marks.length; i++) {
            if (marks[i] > marks[best]) best = i;
        }
        return best;
    }
    public static void main(String[] args) {
        System.out.println(firstLargest(new int[]{70, 95, 95, 82}));
        System.out.println(firstLargest(new int[]{}));
    }
}

Expected result

1
-1
  • Uses > so a tie preserves the first position
  • Handles empty input
  • Does not mutate or sort the input

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