What is the time complexity for the following code?

Question

import java.util.Arrays;
import java.util.Scanner;

class ReverseArray{

static int[] printReverse(int a[],int n){

    Arrays.sort(a);                                   // O(nLogn)
    for(int j = 0 ; j < a.length ; j++){              // O(n)
        System.out.println(a[j]+" ");
    }
    return a;
}

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int a[] = new int[n];

    for(int i = 0 ; i<n ; i++){
        a[i] = sc.nextInt();
    }
    printReverse(a,n);
    sc.close();
}
}

I was trying to practice this code on my own and I wrote this code but I think this is not an optimal code to reverse an array. I have just started my journey in programming so I have only a little idea about Time Complexity. I just want to know the time complexity of this code. The time complexity for this code would be O(N log N)+O(n) right? Correct me if I wrong.

You're correct that a sort is n log n. Because n log n grows faster than n, the complexity is just O(nlogn), you don't add the n. — tgdavies, Jul 24 '21 at 05:37
(Careful. `Arrays.sort` is `O(N log N)` ... but other sorts are not necessarily `O(N log N)`. For example bubble sort and other naive sort algorithms are `O(N^2)`. And so-called counting sort for integers is `O(max(N, R))` where R is `max(inputs) - min(inputs)`. Then there is bogosort :-) ) — Stephen C, Jul 24 '21 at 05:44
Also, this code does NOT reverse the input array, nor prints it in a reverse order. — Nowhere Man, Jul 24 '21 at 05:44
You really think it works? Did you look at the output to see what order it is in? Did you look at the contents of `a` to see what order it is in? — Stephen C, Jul 24 '21 at 06:33

What is the time complexity for the following code?

0 Answers0