AP Computer Science Exam Question

// You are given an array of integers named "nums" and an integer "target". 
// Your task is to write a method "traverseAndManipulate" that will perform the following operations on the array:

// 1. Traverse the array and find all pairs of numbers whose sum equals the target.
// 2. Multiply each of the pairs found in step 1 by a factor of 2.
// 3. Sort the array in ascending order.

// The method should return the modified array as the result. If no pairs are found, or if the array is empty, the method should return an empty array.

// You may assume that the array may contain duplicate numbers, but the same pair should not be counted more than once.

// You may use any additional methods or classes as needed.

// Answer the  following questions and write the code providing detailed explanations for the implementation.

// a) What is the time complexity of your solution?
// b) What is the space complexity of your solution?
// c) Write the code for the "traverseAndManipulate" method.


public class ArrayTraversalManipulation {

    public static int[] traverseAndManipulate(int[] nums, int target) {
        // Step 1: Traverse the array and find all pairs of numbers whose sum equals the target.
        ArrayList<Integer[]> pairs = new ArrayList<>();

        int n = nums.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] + nums[j] == target) {
                    pairs.add(new Integer[]{nums[i], nums[j]});
                }
            }
        }

        // If no pairs found, return empty array
        if (pairs.isEmpty()) {
            return new int[]{};
        }

        // Step 2: Multiply each of the pairs found in step 1 by a factor of 2.
        for (Integer[] pair : pairs) {
            pair[0] *= 2;
            pair[1] *= 2;
        }

        // Step 3: Sort the array in ascending order.
        Arrays.sort(nums);

        //Convert the integer 2D array to a 1D array and return
        int[] result = new int[2 * pairs.size()];
        int index = 0;
        for (Integer[] pair : pairs) {
            result[index++] = pair[0];
            result[index++] = pair[1];
        }

        return result;
    }

    public static void main(String[] args) {
        // Test the traverseAndManipulate method
        int[] nums = {1, 4, 2, 7, 5, 3, 11};
        int target = 8;

        int[] result = traverseAndManipulate(nums, target);

        System.out.println("Modified Array: " + Arrays.toString(result));
    }
}

Explanation:

a) The time complexity of the solution is O(n^2 log n), where n is the length of the given array. The nested for loop runs in O(n^2) time, and the sorting step takes O(n log n) time.

b) The space complexity of the solution is O(n), where n is the length of the given array. We use an ArrayList to store pairs, which can have a maximum of n/2 pairs. The result array also has a maximum size of 2n.

c) The traverseAndManipulate method performs the following steps:

1. Create an ArrayList named pairs to store pairs of numbers whose sum equals the target.
2. Iterate through the array using two nested loops to find all pairs with the given sum.
   • If a pair is found, add it to the pairs list as an Integer array containing the two numbers.
3. If no pairs are found, return an empty array.
4. Multiply each pair in the pairs list by a factor of 2.
5. Sort the original array in ascending order using Arrays.sort() method.
6. Create a result array with a length equal to twice the number of pairs found.
7. Copy each pair from pairs to the result array, converting from the Integer array to individual integers.
8. Return the result array.

In the given sample code, the main method tests the traverseAndManipulate method by providing a sample array nums and a target value of 8. The modified array is printed as the output.