Question:

Consider the following Java implementation of the selection sort algorithm:

public class SelectionSort {
    
    public static void selectionSort(int[] arr) {
        int n = arr.length;
        
        for (int i = 0; i < n-1; i++) {
            int minIndex = i;
            
            for (int j = i+1; j < n; j++) {
                if (arr[j] < arr[minIndex]) {
                    minIndex = j;
                }
            }
            
            int temp = arr[minIndex];
            arr[minIndex] = arr[i];
            arr[i] = temp;
        }
    }
    
    public static void main(String[] args) {
        int[] numbers = {5, 2, 8, 1, 9, 3};
        selectionSort(numbers);
        
        for (int num : numbers) {
            System.out.print(num + " ");
        }
    }
}

Identify the outputs of the selection sort implementation given above.

a) 5 2 8 1 9 3 b) 1 2 3 5 8 9 c) 9 8 5 3 2 1 d) 3 1 2 5 8 9

Explain your answer using step-by-step explanations of the selection sort algorithm.

Answer:

The selection sort algorithm sorts an array by repeatedly finding the minimum element from the unsorted part of the array and swapping it with the element at the beginning of the unsorted part. The process continues until the entire array is sorted.

Let's analyze the given code:

1. The code defines a selectionSort method and a main method.
2. The selectionSort method takes an integer array as input and sorts it in ascending order using the selection sort algorithm.
3. The main method initializes an integer array numbers with values {5, 2, 8, 1, 9, 3}.
4. It calls the selectionSort method, passing the numbers array as an argument.
5. The sorted array elements are printed using a for-each loop.

Now, let's trace the execution of the selection sort algorithm step-by-step:

1. Initial state: numbers = [5, 2, 8, 1, 9, 3].
2. The outer loop iterates from i = 0 to i = n-2 (where n is the size of the array - 6 in this case).
3. In the first iteration (when i = 0):
   • The minimum index minIndex is initially set to 0.
   • The inner loop iterates from j = i+1 to j = n-1, comparing each element with the element at minIndex.
   • The element at index j = 1 (value 2) is found to be smaller than the element at minIndex = 0 (value 5). So, minIndex is updated to 1.
   • The element at index j = 2 (value 8) is found to be greater than the current minimum element 2.
   • The element at index j = 3 (value 1) is found to be smaller than the current minimum element 2. So, minIndex is updated to 3.
   • The element at index j = 4 (value 9) is found to be greater than the current minimum element 1.
   • The element at index j = 5 (value 3) is found to be smaller than the current minimum element 1. So, minIndex is updated to 5.
   • The inner loop ends.
   • Swap the element at minIndex = 5 (value 3) with the element at i = 0 (value 5).
   • The array becomes: [3, 2, 8, 1, 9, 5].
4. In the second iteration (when i = 1):
   • The minimum index minIndex is initially set to 1.
   • The inner loop iterates from j = i+1 to j = n-1, comparing each element with the element at minIndex.
   • The element at index j = 2 (value 8) is found to be greater than the current minimum element 2.
   • The element at index j = 3 (value 1) is found to be smaller than the current minimum element 2. So, minIndex is updated to 3.
   • The element at index j = 4 (value 9) is found to be greater than the current minimum element 1.
   • The inner loop ends.
   • Swap the element at minIndex = 3 (value 1) with the element at i = 1 (value 2).
   • The array becomes: [3, 1, 8, 2, 9, 5].
5. The remaining iterations follow the same steps.
6. After the final iteration (when i = 4):
   • The array becomes: [1, 2, 3, 5, 8, 9].

Therefore, the output of the given code will be Option B: 1 2 3 5 8 9.