Sort an almost sorted array where only two elements where swapped

Given an array where only two elements where swapped , sort the array with O(n) time complexity and one swap

 
Input:  arr[] = {10, 20, 60, 40, 50, 30}  
// 30 and 60 are swapped
Output: arr[] = {10, 20, 30, 40, 50, 60}

Input:  arr[] = {10, 20, 40, 30, 50, 60}  
// 30 and 40 are swapped
Output: arr[] = {10, 20, 30, 40, 50, 60}

Input:   arr[] = {1, 5, 3}
// 3 and 5 are swapped
Output:  arr[] = {1, 3, 5}

 Concept behind:

   If you swap two elements in a sorted(ascending) array, the first out of order element will be greater than its next element and the next out of order element will be less than its previous element.

 Consider:  {10, 20, 40, 30, 50, 60} 

 Here the first out of order element is 40( 40 > 30) -> a[i] > a[i+1]

 The next out of order element is  30 (30 < 40) -> a[i] < a[i-1]

 Hence we will first find the first out of order element and from the next element we check for the other out of order element using the above conditions.

 Code:

 import java.util.*;

 public class Solution {
    
    public static void main(String[] args) {
        int[] a = {10, 20, 40, 30, 50, 60} ;
        int n = a.length;
        int index1 = 0;
        int index2 = n-1;
       
        for(int i=0; i<n-1; i++){
            if(a[i]>a[i+1]) {
                index1 = i;
                for(int j=i+1; j<n; j++){
                    if(a[j]<a[j-1]) {
                        index2 = j;
                        break;
                    }
                }
                break;
            }
         }
        
        int temp = a[index1];
        a[index1] = a[index2];
        a[index2] = temp;
        
       for(int i=0; i<n; i++){
          System.out.print(a[i]+" ");
       }
       
     
    }
}

 Output:

 
10 20 30 40 50 60



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