Count of Numbers such that difference between the number and sum of its digits not less than L in C++

C++Server Side ProgrammingProgramming

We are given a number N and another number L. The goal is to find the numbers between 1 and N that have a difference between the number itself and the sum of its digits is not less than L.

If N=23, L=10 then the count of such numbers will be 4.

23-(2+3)=18, 22-(2+2)=18, 21-(2+1)=18, 20-(2+0)=18.

All above numbers meet the condition

But 19-(1+9)=9 which is less than L, similarly 18,17….1.

Let us understand with examples

Input − N=30 L=19

Output − Count of Numbers such that difference between the number and sum of its digits not less than L are − 1

Explanation − Only 30 meets the condition, 30-(3+0)=27 > 19

Input − N=123330 L=5466

Output − Count of Numbers such that difference between the number and sum of its digits not less than L are − 6841

Approach used in the below program is as follows

Using binary search we will find the first number that meets the condition. If that number is num then the condition will also be true for num+1 and so on.

If any current mid value satisfies the condition then all numbers between mid and end will also satisfy this condition so we can simply add end-mid+1 to count.

  • Take num and L as long variables.

  • Function Digit_sum(LL num) takes a number num and returns the sum of its digits.

  • Take initial sum as total=0.

  • Using a while loop, add reminder num%10 to total and reduce num by 10. Do this until num>0.

  • Return total as sum of digits of num.

  • Function Less_than_L(LL num, LL L) takes a number num and a number L and returns count of Numbers such that difference between the number and sum of its digits not less than L

  • Take the initial count as 0.

  • Implement binary search using while loop where start=1 and end=num.

  • Calculate middle number as temp=(start+end)/2.

  • If the difference between temp and sum of digits of temp is not less than L then all numbers greater than temp will also satisfy the same condition.

  • Count of such numbers including temp will be num-temp+1. Add this to count. And set end=temp-1.

  • Otherwise set start=temp+1.

  • At the end of binary search count will have numbers with difference between them and sum of digits not less than L

  • Return count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int Digit_sum(LL num){
   LL total = 0;
   while (num > 0){
      total += num % 10;
      num = num/10;
      z}
   return total;
}
LL Less_than_L(LL num, LL L){
   LL count = 0;
   LL start = 1;
   LL end = num;
   while (start <= end){
      LL temp = (end + start) / 2;
      LL temp_2 = temp - Digit_sum(temp);
      if (temp_2 >= L){
         count = num - temp + 1;
         end = temp - 1;
      }
      else{
         start = temp + 1;
      }
   }
   return count;
}
int main(){
   LL num = 234516;
   LL L = 235;
   cout<<"Count of Numbers such that difference between the number and sum of its digits not
   less than L are: "<< Less_than_L(num, L);
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of Numbers such that difference between the number and sum of its digits not less than L are: 234267
raja
Published on 03-Dec-2020 07:24:38
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