Count of Digits Inverted in Their Position

The program must accept an integer N as the input. The task is to count and print the number of digits in N whose position (from the rightmost side) is the same as the digit itself. If no such digits are found, print -1.

Note: Position starts from 1 from the rightmost side.

Boundary Condition(s):
1 <= N <= 10^9

Input Format:
The first line contains the integer N.

Output Format:
The first line contains either the count of the digits whose position (from the rightmost side) is the same as the digit itself or -1.

Example Input/Output 1:
Input:
3412

Output:
2

Explanation:
The digit 1 is in the 1st position (from the rightmost side) and the digit 2 is in the 2nd position. Hence the count is 2.

Example Input/Output 2:
Input:
4567

Output:
-1

Explanation:
There are no digits whose position (from the rightmost side) is the same as the digit itself.

n = input().strip()
count = sum(1 for idx, digit in enumerate(reversed(n), 1) if int(digit) == idx)
print(count if count > 0 else -1)

#include<stdio.h>
#include<string.h>

int main() {
    char n[12];
    scanf("%s", n);
    int len = strlen(n);
    int count = 0;
    for(int i = 0; i < len; i++) {
        if(n[len - i - 1] - '0' == i + 1) count++;
    }
    printf("%d", (count > 0 ? count : -1));
    return 0;
}

#include <iostream>
#include <string>
using namespace std;

int main() {
    string n;
    cin >> n;
    int count = 0;
    for(int i = 0; i < n.size(); i++) {
        if(n[n.size() - i - 1] - '0' == i + 1) count++;
    }
    cout << (count > 0 ? count : -1);
    return 0;
}

import java.util.Scanner;

public class CountInvertedDigits {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String n = sc.next();
        int count = 0;
        for(int i = 0; i < n.length(); i++) {
            if(n.charAt(n.length() - i - 1) - '0' == i + 1) count++;
        }
        System.out.println(count > 0 ? count : -1);
    }
}