Number of Digit One in C++

Suppose we have a number n, We have to count the total number of digit 1 appearing in all non-negative numbers less than or equal to n. So if the input is 15, then the output will be 8, because the numbers containing 1 is [1,10,11,12,13,14,15], there are 8 1s.

To solve this, we will follow these steps −

• ret := 0

• for initializing i := 1, when i <= n, i = i * 10 do −

  • a := n / i, b := n mod i, x := a mod 10

  • if x is same as 1, then,

    • ret = ret + (a / 10) * i + (b + 1)

  • Otherwise when x is same as 0, then −

    • ret = ret + (a / 10) * i

  • Otherwise

    • ret = ret + (a / 10 + 1) * i

• return ret

Example

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int countDigitOne(int n) {
      int ret = 0;
      for(long long int i = 1; i <= n; i*= (long long int)10){
         int a = n / i;
         int b = n % i;
         int x = a % 10;
         if(x ==1){
            ret += (a / 10) * i + (b + 1);
         }
         else if(x == 0){
            ret += (a / 10) * i;
         } else {
            ret += (a / 10 + 1) *i;
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   cout << (ob.countDigitOne(15));
}

Input

15

Output

8