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.


diffFirstLastDigits(N, X)

   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
   first := len
   return |first - last|


 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;
   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);


Enter number and number of digits from first and last: 568424 2 
Difference: 32

Updated on: 30-Jul-2019


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