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

Output

We can remove 2 to make it divisible by 3.

Input

Output

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.

Hafeezul Kareem

Updated on: 26-Oct-2021

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