Number of digits to be removed to make a number divisible by 3 in C++

You are given a number in string. You need to find how many digits need to be removed to make it divisible by 3.

We make a number divisible by removing at most 2 digits. So, the maximum number of digits to be removed to make it divisible by 3 is 2.

Let's see some examples.

Input

92

Output

1

We can remove 2 to make it divisible by 3.

Input

999

Output

0

The given number itself is divisible by 3.

Algorithm

• Initialise the number in string.

• Find the sum of the number.

• If the sum is divisible by 3, then return 0.

• If the sum is not divisible by 3 and the length of the number is 1, then we can't make it divisible by 3. Return -1.

• Iterate over the number.

  • Remove one digit from the number and check the divisibility.

  • If the above condition satisfies, then return 1.

• Check the length of the number again. If the length is 2, then return -1.

• Else return 2.

Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
int getNumSum(string n) {
   int len = n.length(), sum = 0;
   for (int i = 0; i < len; i++) {
      sum += (int)n[i];
   }
   return sum;
}
int getDigitsCount(string num) {
   int n = num.length();
   int sum = getNumSum(num);
   if (sum % 3 == 0) {
      return 0;
   }
   if (n == 1) {
      return -1;
   }
   for (int i = 0; i < n; i++) {
      int currentDigit = num[i] - '0';
      if (sum % 3 == currentDigit % 3) {
         return 1;
      }
   }
   if (n == 2) {
      return -1;
   }
   return 2;
}
int main() {
   string num = "7536836";
   cout << getDigitsCount(num) << endl;
   return 0;
}

Output

If you run the above code, then you will get the following result.

1