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** Link for the Problem** – Day 20: Sorting – Hacker Rank Solution

Day 20: Sorting – Hacker Rank Solution

**Problem:**

**Objective**

Today, we’re discussing a simple sorting algorithm called *Bubble Sort*. Check out the Tutorial tab for learning materials and an instructional video!

Consider the following version of Bubble Sort:

for (int i = 0; i < n; i++) { // Track number of elements swapped during a single array traversal int numberOfSwaps = 0; for (int j = 0; j < n - 1; j++) { // Swap adjacent elements if they are in decreasing order if (a[j] > a[j + 1]) { swap(a[j], a[j + 1]); numberOfSwaps++; } } // If no elements were swapped during a traversal, array is sorted if (numberOfSwaps == 0) { break; } }

**Task**

Given an array, , of size distinct elements, sort the array in *ascending* order using the *Bubble Sort* algorithm above. Once sorted, print the following lines:

`Array is sorted in numSwaps swaps.`

where is the number of swaps that took place.`First Element: firstElement`

where is the*first*element in the sorted array.`Last Element: lastElement`

where is the*last*element in the sorted array.

**Hint:** To complete this challenge, you will need to add a variable that keeps a running tally of *all* swaps that occur during execution.

**Example**

original a: 4 3 1 2 round 1 a: 3 1 2 4 swaps this round: 3 round 2 a: 1 2 3 4 swaps this round: 2 round 3 a: 1 2 3 4 swaps this round: 0

In the first round, the is swapped at each of the comparisons, ending in the last position. In the second round, the is swapped at of the comparisons. Finally, in the third round, no swaps are made so the iterations stop. The output is the following:

Array is sorted in 5 swaps. First Element: 1 Last Element: 4

**Input Format**

The first line contains an integer, , the number of elements in array .

The second line contains space-separated integers that describe .

**Constraints**

- , where .

**Output Format**

Print the following three lines of output:

`Array is sorted in numSwaps swaps.`

where is the number of swaps that took place.`First Element: firstElement`

where is the*first*element in the sorted array.`Last Element: lastElement`

where is the*last*element in the sorted array.

**Sample Input 0**

3 1 2 3

**Sample Output 0**

Array is sorted in 0 swaps. First Element: 1 Last Element: 3

**Explanation 0**

The array is already sorted, so swaps take place and we print the necessary lines of output shown above.

**Sample Input 1**

3 3 2 1

**Sample Output 1**

Array is sorted in 3 swaps. First Element: 1 Last Element: 3

**Explanation 1**

The array is *not sorted*, so we perform the following swaps. Each line shows after each single element is swapped.

After swaps, the array is sorted.

Day 20: Sorting – Hacker Rank Solution

import java.util.Scanner; /** * @author Techno-RJ * */ public class Day20Sorting { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int a[] = new int[n]; for (int a_i = 0; a_i < n; a_i++) { a[a_i] = in.nextInt(); } in.close(); int count = 0; boolean swapped = true; for (int j = a.length - 1; (j >= 0 && swapped); j--) { swapped = false; for (int k = 0; k < j; k++) { if (a[k] > a[k + 1]) { int temp = a[k]; count++; a[k] = a[k + 1]; a[k + 1] = temp; swapped = true; } } } System.out.println("Array is sorted in " + count + " swaps."); System.out.println("First Element: " + a[0]); System.out.println("Last Element: " + a[a.length - 1]); } }

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