# Absolute difference between the first X and last X Digits of N?

Here we will see how to get the differences between first and last X digits of a number N. The number and X are given. To solve this problem, we have to find the length of the number, then cut the last x digits using modulus operator. After that cut all digits from the number except first x digits. Then get the difference, and return the result. Let the number is N = 568424. The X is 2 so first two digits are 56, and last two digits are 24. The difference is (56 - 24) = 32.

## Algorithm

#### diffFirstLastDigits(N, X)

begin
p := 10^X
last := N mod p
len := length of the number N
while len is not same as X, do
N := N / 10
len := len -1
done
first := len
return |first - last|
end

## Example

Live Demo

#include <iostream>
#include <cmath>
using namespace std;
int lengthCount(int n){
return floor(log10(n) + 1);
}
int diffFirstLastDigits(int n, int x) {
int first, last, p, len;
p = pow(10, x);
last = n % p;
len = lengthCount(n);
while(len != x){
n /= 10;
len--;
}
first = n;
return abs(first - last);
}
main() {
int n, x;
cout << "Enter number and number of digits from first and last: ";
cin >> n >> x;
cout << "Difference: " << diffFirstLastDigits(n,x);
}

## Output

Enter number and number of digits from first and last: 568424 2
Difference: 32
Published on 26-Jul-2019 06:45:54