Alphabets Positions Reversed

The program must accept N alphabets as the input. The program must print the alphabet position of each alphabet in reverse order as the output.

Boundary Condition(s):
1 <= N <= 10^5

Input Format:
The first line contains the integer N.
The second line contains N characters separated by space(s).

Output Format:
The first line contains the alphabet position of N alphabets in the reverse order.

Example Input/Output 1:
Input:
5
j a e E b

Output:
2 5 5 1 10

Explanation:
The alphabet position of j is 10 in the alphabetical order.
The alphabet position of a is 1 in the alphabetical order.
The alphabet position of e is 5 in the alphabetical order.
The alphabet position of E is 5 in the alphabetical order.
The alphabet position of b is 2 in the alphabetical order.
In the reverse order, the alphabets position are 2 5 5 1 10
Hence the output is 2 5 5 1 10

Example Input/Output 2:
Input:
4
L f V d

Output:
4 22 6 12 

#include <stdio.h>
#include <ctype.h>

int main() {
    int n;
    scanf("%d", &n);
    char arr[n];

    for(int i = 0; i < n; i++) {
        scanf(" %c", &arr[i]);
    }

    for(int i = n - 1; i >= 0; i--) {
        printf("%d ", tolower(arr[i]) - 'a' + 1);
    }

    return 0;
}

n = int(input())
alphabets = list(input().split())

positions = [str(ord(ch.lower()) - ord('a') + 1) for ch in alphabets]
print(' '.join(positions[::-1]))

import java.util.Scanner;

public class AlphabetPosition {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        char[] arr = new char[n];

        for(int i = 0; i < n; i++) {
            arr[i] = scanner.next().charAt(0);
        }

        for(int i = n - 1; i >= 0; i--) {
            System.out.print(Character.toLowerCase(arr[i]) - 'a' + 1 + " ");
        }
    }
}

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    char arr[n];

    for(int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    for(int i = n - 1; i >= 0; i--) {
        cout << tolower(arr[i]) - 'a' + 1 << " ";
    }

    return 0;
}