Count unique numbers that can be generated from N by adding one and removing trailing zeros in C++

We are given a number N as input. Perform two operations on N and identify the count of unique numbers generated in the process. Steps will −

• Add 1 to number

• Remove trailing zeros from the generated number, if any

If N is 8 then numbers generated will be

Applying step 1− 8 → 9 →

Applying step 2− 1 → ( removed 0 from 10 )

Applying step 1: 2 → 3 → 4 → 5 → 6 → 7 → 8 ( same sequence )

Count of unique numbers will be 9.

For Example

Input

N=21

Output

Count of unique numbers that can be generated from N by adding one and
removing trailing zeros are: 18

Explanation

Numbers will be: 21, 22, 23, 24, 25, 26, 27, 28, 29, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3 −−−now same sequence Unique numbers are: 18

Input

N=38

Output

Count of unique numbers that can be generated from N by adding one and removing trailing zeros are: 11

Explanation

Numbers will be:
38, 39, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4 −−−now same sequence
Unique numbers are: 11

Approach used in the below program is as follows −

In this approach we will create an unordered set which will contain all unique numbers generated after applying steps 1 and 2. In case the numbers repeat, then we will stop the iteration. The size of the set will give us the count of unique numbers generated in the process.

• Take the number N as integer.

• Take the unordered_set<int> U_S for inserting generated numbers.

• Function unique_N(unordered_set<int>& U_S, int N) takes the set and N and adds numbers to the set U_S until all numbers are unique in it.

• If U_S.count(N) returns 1 that means N already exists in the set. So the numbers would repeat, return from the function.

• Otherwise insert N to the set and apply operation 1 ( increment by 1).

• Check if number N has trailing zeroes (is multiple of 10).

• If N % 10 is 0, then remove trailing zero by dividing it by 10.

• Call function unique_N() with updated N.

• After returning from the function, take count as the size of the set U_S.

• Print result as count.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void unique_N(unordered_set<int>& U_S, int N){
   if (U_S.count(N)){
      return;
   }
   U_S.insert(N);
   N = N + 1;
   while (N % 10 == 0){
      N = N / 10;
   }
   unique_N(U_S, N);
}
int main(){
   int N = 7;
   unordered_set<int> U_S;
   unique_N(U_S, N);
   int count = U_S.size();
   cout<<"Count of unique numbers that can be generated from N by adding one and removing
      trailing zeros are: "<<count;
   return 0;
}

Output

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

Count of unique numbers that can be generated from N by adding one and removing trailing zeros are: 9