Count of different numbers divisible by 3 that can be obtained by changing at most one digit

Given a string str representing a number having N digits, the task is to calculate the number of ways to make the given number divisible by 3 by changing at most one digit of the number.

Examples:

Input: str[] = “23”
Output: 7
Explanation: Below are the numbers that can be made from the string which are divisible by 3 – 03, 21, 24, 27, 33, 63, 93
1.Change 2 to 0 (0+3)=3 divisible by 3
2.Change 3 to 1 (2+1)=3 divisible by 3
3 change 3 to 4 (2+4)=6 divisible by 3
4 change 2 to 3 sum is 6 divisible by 3
Similarly there are total 7 number of ways to make the given number divisible by 3

Input: str[] = “235”
Output: 9

Approach: The idea is very simple to solve this problem. Calculate the sum of digits of a given number and then for each index, remove that digit and try all possible digits from 0 to 9 and see if the sum is divisible by 3 or not. Follow the steps below to solve the problem:

• Initialize the variable sum as 0 to store the sum of digits of the number.
• Iterate over a range [0, N] using the variable i and perform the following steps:
  • Add the value of digit at i-th index in the variable sum.
• Initialize the variable count as 0 to store the answer.
• If the number itself is divisible by 3 then increment count by one.
• Iterate over a range [0, N] using the variable i and perform the following steps:
  • Initialize the variable remaining_sum as sum-(number.charAt(i)-48).
  • Iterate over a range [0, 9] using the variable j and perform the following steps:
    • Add the value of j to the variable remaining_sum and if remaining_sum is divisible by 3 and j is not equal to the digit at i-th index, then add the value of count by 1.
• After performing the above steps, print the value of count as the answer.

Below is the implementation of the above approach..

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// possible numbers divisible by 3
void findCount(string number)
{

    // Calculate the sum
    int sum = 0;
    for (int i = 0; i < number.length(); ++i) {
        sum += number[i] - 48;
    }

    // Store the answer
    int count = 0;

    // Consider the edge case when
    // the number itself is divisible by 3
    // The count will be added by 1
    if (sum % 3 == 0)
        count++;

    // Iterate over the range
    for (int i = 0; i < number.length(); ++i) {

        // Decreasing the sum
        int remaining_sum = sum - (number[i] - 48);

        // Iterate over the range
        for (int j = 0; j <= 9; ++j) {

            // Checking if the new sum
            // is divisible by 3 or not
            if ((remaining_sum + j) % 3 == 0
                && j != number[i] - 48) {

                // If yes increment
                // the value of the count
                ++count;
            }
        }
    }
    cout << count;
}

// Driver Code
int main()
{
    // Given number
    string number = "235";

    findCount(number);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG {

    // Function to count the number of
    // possible numbers divisible by 3
    public static void findCount(String number)
    {

        // Calculate the sum
        int sum = 0;
        for (int i = 0; i < number.length(); ++i) {
            sum += number.charAt(i) - 48;
        }

        // Store the answer
        int count = 0;
        if (sum % 3 == 0)
            count++;

        // Iterate over the range
        for (int i = 0; i < number.length(); ++i) {

            // Decreasing the sum
            int remaining_sum
                = sum - (number.charAt(i) - 48);

            // Iterate over the range
            for (int j = 0; j <= 9; ++j) {

                // Checking if the new sum
                // is divisible by 3 or not
                if ((remaining_sum + j) % 3 == 0
                    && j != number.charAt(i) - 48) {

                    // If yes increment
                    // the value of the count
                    ++count;
                }
            }
        }
        System.out.println(count);
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given number
        String number = "235";

        findCount(number);
    }
}

Python3

# Python program for the above approach

# Function to count the number of
# possible numbers divisible by 3


def findCount(number):

    # Calculate the sum
    sum = 0
    for i in range(len(number)):
        sum += int(number[i]) - 48

    # Store the answer
    count = 0
    if(sum % 3 == 0):
      count += 1

    # Iterate over the range
    for i in range(len(number)):

        # Decreasing the sum
        remaining_sum = sum - (int(number[i]) - 48)

        # Iterate over the range
        for j in range(10):

            # Checking if the new sum
            # is divisible by 3 or not
            if ((remaining_sum + j) % 3 == 0 and j != int(number[i]) - 48):

                # If yes increment
                # the value of the count
                count += 1

    print(count)

# Driver Code


# Given number
number = "235"
findCount(number)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG {

    // Function to count the number of
    // possible numbers divisible by 3
    static void findCount(string number)
    {

        // Calculate the sum
        int sum = 0;
        for (int i = 0; i < number.Length; ++i) {
            sum += (int)number[i] - 48;
        }

        // Store the answer
        int count = 0;
        if (sum % 3 == 0)
            count++;

        // Iterate over the range
        for (int i = 0; i < number.Length; ++i) {

            // Decreasing the sum
            int remaining_sum = sum - ((int)number[i] - 48);

            // Iterate over the range
            for (int j = 0; j <= 9; ++j) {

                // Checking if the new sum
                // is divisible by 3 or not
                if ((remaining_sum + j) % 3 == 0
                    && j != number[i] - 48) {

                    // If yes increment
                    // the value of the count
                    ++count;
                }
            }
        }
        Console.Write(count);
    }

    // Driver Code
    public static void Main()
    {

        // Given number
        string number = "235";

        findCount(number);
    }
}

Javascript

<script>
// Javascript program for the above approach

// Function to count the number of
// possible numbers divisible by 3
function findCount(number) {
  // Calculate the sum
  let sum = 0;
  for (let i = 0; i < number.length; ++i) {
    sum += number[i] - 48;
  }

  // Store the answer
  let count = 0;
  if(sum % 3 == 0) count++;
  // Iterate over the range
  for (let i = 0; i < number.length; ++i) {
    // Decreasing the sum
    let remaining_sum = sum - (number[i] - 48);

    // Iterate over the range
    for (let j = 0; j <= 9; ++j) {
      // Checking if the new sum
      // is divisible by 3 or not
      if ((remaining_sum + j) % 3 == 0 && j != number[i] - 48) {
        // If yes increment
        // the value of the count
        ++count;
      }
    }
  }
  document.write(count);
}

// Driver Code

// Given number
let number = "235";

findCount(number);

// This code is contributed by gfgking

</script>

Output

9

Time Complexity: O(N)
Auxiliary Space: O(1)