AP CSA Array & ArrayList Mini FRQs - Write Real Code (Interactive)

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AP CSA Array & ArrayList Mini FRQs

8 exam-level problems. Write your solution, run it, and get instant auto-graded feedback. This is how you build FRQ confidence.

AP CSA Exam: May 15, 2026. Reading about arrays is not enough. You need to write the code yourself.
How it works: Each problem gives you a method signature. Write your implementation in the editor. Click Run & Grade to execute your code against hidden test cases. You will see a pass or fail result with your output vs expected output. Click Show Solution if you get stuck. Your code runs in a live Java environment — real compilation, real execution.
Your Score 0 / 8 passed

Remove All Occurrences

Write a method removeAll that takes an ArrayList and an int target, and removes ALL occurrences of target from the list. The method modifies the list in place and returns nothing.

Hint: Traverse backwards to avoid index-shifting bugs.
Solution: 
public static void removeAll(ArrayList list, int target)
{
    for (int i = list.size() - 1; i >= 0; i--)
    {
        if (list.get(i) == target)
        {
            list.remove(i);
        }
    }
}

Remove Duplicates

Write a method removeDuplicates that takes an ArrayList and removes duplicate values, keeping only the first occurrence of each. Modifies the list in place.

Hint: For each element, check if it appeared earlier in the list.
Solution: 
public static void removeDuplicates(ArrayList list)
{
    for (int i = list.size() - 1; i >= 1; i--)
    {
        for (int j = 0; j < i; j++)
        {
            if (list.get(i).equals(list.get(j)))
            {
                list.remove(i);
                break;
            }
        }
    }
}

Shift Left

Write a method shiftLeft that takes an int[] and shifts every element one position to the left. The first element wraps around to the last position.

Hint: Save the first element, shift everything left, then put the saved value at the end.
Solution: 
public static void shiftLeft(int[] arr)
{
    if (arr.length <= 1)
    {
        return;
    }
    int first = arr[0];
    for (int i = 0; i < arr.length - 1; i++)
    {
        arr[i] = arr[i + 1];
    }
    arr[arr.length - 1] = first;
}

Reverse Array In Place

Write a method reverse that reverses an int[] in place. Do not create a new array.

Hint: Swap elements from the outside in, using two index variables moving toward the center.
Solution: 
public static void reverse(int[] arr)
{
    int left = 0;
    int right = arr.length - 1;
    while (left < right)
    {
        int temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

Filter to New List

Write a method filterAbove that takes an int[] and an int threshold, and returns a new ArrayList containing only elements greater than threshold.

Hint: Create a new ArrayList, loop through the array, add elements that pass the condition.
Solution: 
public static ArrayList filterAbove(int[] arr, int threshold)
{
    ArrayList result = new ArrayList();
    for (int i = 0; i < arr.length; i++)
    {
        if (arr[i] > threshold)
        {
            result.add(arr[i]);
        }
    }
    return result;
}

Count Long Strings

Write a method countLong that takes a String[] and an int minLen, and returns the number of Strings whose length is greater than or equal to minLen.

Hint: Standard accumulator pattern with a length condition.
Solution: 
public static int countLong(String[] words, int minLen)
{
    int count = 0;
    for (int i = 0; i < words.length; i++)
    {
        if (words[i].length() >= minLen)
        {
            count++;
        }
    }
    return count;
}

Swap Adjacent Pairs

Write a method swapPairs that takes an int[] and swaps every pair of adjacent elements. If the array has odd length, the last element stays in place.

Hint: Loop with step 2. Swap arr[i] and arr[i+1].
Solution: 
public static void swapPairs(int[] arr)
{
    for (int i = 0; i < arr.length - 1; i += 2)
    {
        int temp = arr[i];
        arr[i] = arr[i + 1];
        arr[i + 1] = temp;
    }
}

Row Sums (2D Array)

Write a method rowSums that takes a int[][] and returns a new int[] where each element is the sum of the corresponding row.

Hint: Create a result array with grid.length elements. Nested loop to sum each row.
Solution: 
public static int[] rowSums(int[][] grid)
{
    int[] sums = new int[grid.length];
    for (int r = 0; r < grid.length; r++)
    {
        int total = 0;
        for (int c = 0; c < grid[r].length; c++)
        {
            total += grid[r][c];
        }
        sums[r] = total;
    }
    return sums;
}
Final Score 0 / 8 passed

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