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

C++Server Side ProgrammingProgramming

In this tutorial, we are going to write a program that finds the number of digits to be removed to make a number divisible by 3.

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. The maximum number of digits to be removed to make it divisible by 3 is 2.,

Let's see the steps to solve the problem.

• 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.

## Example

Let's see the code.

Live Demo

#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

## Conclusion

If you have any queries in the tutorial, mention them in the comment section.

Published on 03-Jul-2021 05:04:03