📌How to reverse an array?📌

Question: Write a program to reverse an array or string.

Example:

Input  : arr[] = {1, 2, 3}
Output : ar[] = {3, 2, 1}

Logic: O ( n )

• Take two pointers i and j
• put i on arr [ 0 ] and put j on arr [ arr.length -1 ]
• swap i and j till i becomes greater than or equal to j

import java.util.Scanner;

public class Solution{

//This function depicts the recursive solution of the question.
    public static void reverseArray(int arr[],int i, int j){
                if(i >= j){
                    return;             
                }

            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
            reverseArray(arr,i++,j--);
    }

    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int arr[] = new int[n];
        for(int i=0 ; i<arr.length ; i++){
            arr[i] = sc.nextInt();
        }
        int i=0;
        int j = arr.length-1;
        reverseArray(arr, i, j);//For recurssive function above

            //This is itterative method.
        while(i<j){
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
            i++;
            j--;
        }

        for(int k=0; k<arr.length; k++){
            System.out.print(arr[k]+" ");
        }
        System.out.println();
    }
}

Comments

[Josh Melo]

Nice post! Question though, wouldn't the time complexity be O(n / 2 ) since it only goes through half the length of the array?

[Kaiwalya Koparkar]

That's true but if you simplify (n/2) so it becomes (n^-2). As power is negative we neglect it. and hence the total complexity becomes O(n). What are your views?