Algorithm Steps
- If you are at the start of the array then go to the right element (from arr[0] to arr[1]).
- If the current array element is larger or equal to the previous array element then go one step right
if (arr[i] >= arr[i-1])
i++;
- If the current array element is smaller than the previous array element then swap these two elements and go one step backwards
if (arr[i] < arr[i-1])
{
swap(arr[i], arr[i-1]);
i--;
}
- Repeat steps 2) and 3) till ‘i’ reaches the end of the array (i.e- ‘n-1’)
- If the end of the array is reached then stop and the array is sorted.
Java
// Java Program to implement Gnome Sort Â
import java.util.Arrays; public class GFG { Â static void gnomeSort(int arr[], int n) Â { Â Â int index = 0; Â
  while (index < n) {    if (index == 0)     index++;    if (arr[index] >= arr[index - 1])     index++;    else {     int temp = 0;     temp = arr[index];     arr[index] = arr[index - 1];     arr[index - 1] = temp;     index--;    }   }   return;  } Â
 // Driver program to test above functions.  public static void main(String[] args)  {   int arr[] = { 34, 2, 10, -9 }; Â
  gnomeSort(arr, arr.length); Â
  System.out.print("Sorted sequence after applying Gnome sort: ");   System.out.println(Arrays.toString(arr));  } } Â
// Code Contributed by Mohit Gupta_OMG |
Output:
Sorted sequence after applying Gnome sort: [-9, 2, 10, 34]
Time Complexity: O(n2). As there is no nested loop (only one while) it may seem that this is a linear O(n) time algorithm. But the time complexity is O(n^2) as the variable ‘index’ in the program doesn’t always get incremented, it gets decremented too.Â
Auxiliary Space: O(n)
For understanding of Arrays.toString(), Please refer : https://www.geeksforgeeks.org/arrays-tostring-in-java-with-examples/ Please refer complete article on Gnome Sort for more details!
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