Largest integer divisible by 9 after inserting any digit in N

Last Updated : 16 Jun, 2022

Given a large integer N, the task is to find the largest possible integer which is divisible by 9 after inserting exactly one digit from 0 to 9 anywhere in the N.

Note: Leading zeroes are not allowed.

Examples:

Input: N = "12"
Output: 612
Explanation: The numbers which are divisible by 9 after inserting digit are 612, 162, 126.
And the largest integer is 612. So the output is 612

Input: N = "1346182944512412414214"
Output: 81346182944512412414214

Approach: The problem can be solved with the help of the below observation:

The idea is to insert the digit 0 to 9 at every position in the N and check is it divisible by 9 (divisible only if sum of digits is divisible by 9).
If it is found to be true then store the maximum value with position where digit is inserted and finally print the maximum value.

Follow the below steps to solve the problem:

Initialize a pair variable, ( say maxi = {"", 0}) to store the largest integer which is divisible by 9 with the position where the digit is inserted.
Iterate over the range [0, len) and calculate the digit sum of the number (say stored in variable sum).
Iterate over the range [0, len) and perform the following steps:
- Iterate for ch = '0' to '9' and perform the following steps:
  - Check for leading 0s, if it is found to be true then continue with the iteration.
  - Check if (ch - '0') + sum is divisible by 9 (i.e. sum after inserting that digit in N). If it is found to be true then set or update the value of maxi to the new value.
Finally, print the value of the maximum number (stored in maxi.first).

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check number is divisible by 9
bool isDivBy9(char c, int sum)
{
    return (((c - '0') + sum) % 9 == 0);
}

// Function to find the largest number
pair<string, int> maxNumber(pair<string,
                                 int>
                                s1,
                            string s2, int i)
{
    if ((s1.first[s1.second] - '0') > (s2[s1.second] - '0'))
        return s1;
    return { s2, i };
}

// Function to find the largest integer
// which is divisible by 9 after
// inserting any digit in the N
string findLargestNumber(string N, int len)
{
    // Stores the largest integer which is
    // divisible by 9
    pair<string, int> maxi = { "", 0 };

    // Stores the sum of digits of N
    int sum = 0;
    for (int i = 0; i < len; i++) {

        // Update the value of sum
        sum += (N[i] - '0');
    }

    for (int i = 0; i <= len; i++) {
        for (char ch = '0'; ch <= '9'; ch++) {

            // Skip leading zeroes
            if (i == 0 && ch == '0')
                continue;

            // Check number is divisible by 9
            if (isDivBy9(ch, sum)) {

                // Check maxi is not set
                if (maxi.first.length() == 0) {

                    // Set value of maxi
                    maxi = {
                        N.substr(0, i) + ch + N.substr(i), i
                    };
                }
                else {

                    // Update value of maxi
                    maxi = maxNumber(maxi,
                                     N.substr(0, i) + ch + N.substr(i), i);
                }
            }
        }
    }

    // Print the value of maxi.first
    return maxi.first;
}

// Driver Code
int main()
{
    string N = "12";
    int len = N.length();

    // Function call
    cout << findLargestNumber(N, len);
    return 0;
}

Java

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

class GFG {

  // Function to check number is divisible by 9
  static boolean isDivBy9(char c, int sum)
  {
    return (((c - '0') + sum) % 9 == 0);
  }

  // Function to find the largest number
  static String[] maxNumber(String[] s1, String s2, int i)
  {

    if ((s1[0].charAt(Integer.parseInt(s1[1])) - '0')
        > (s2.charAt(Integer.parseInt(s1[1])) - '0'))
      return s1;
    String[] map = { s2, Integer.toString(i) };
    return map;
  }

  // Function to find the largest integer
  // which is divisible by 9 after
  // inserting any digit in the N
  static String findLargestNumber(String N, int len)
  {
    
    // Stores the largest integer which is
    // divisible by 9
    String[] maxi = { "", "0" };

    // Stores the sum of digits of N
    int sum = 0;
    for (int i = 0; i < len; i++) {

      // Update the value of sum
      sum += (N.charAt(i) - '0');
    }

    for (int i = 0; i <= len; i++) {
      for (char ch = '0'; ch <= '9'; ch++) {

        // Skip leading zeroes
        if ((i == 0) && (ch == '0'))
          continue;

        // Check number is divisible by 9
        if (isDivBy9(ch, sum)) {

          // Check maxi is not set
          if ((maxi[0]) == "") {

            // Set value of maxi
            maxi[0] = N.substring(0, i) + ch
              + N.substring(i);
            maxi[1] = Integer.toString(i);
          }
          else {

            // Update value of maxi
            maxi = maxNumber(
              maxi,
              N.substring(0, i) + ch
              + N.substring(i),
              i);
          }
        }
      }
    }

    // Print the value of maxi.first
    return maxi[0];
  }

  // driver code
  public static void main(String[] args)
  {
    String N = "12";
    int len = N.length();

    // Function call
    System.out.print(findLargestNumber(N, len));
  }
}

// This code is contributed by phasing17

Python3

# Python3 program for the above approach

# Function to check number is divisible by 9
def isDivBy9(c, sums):
    return ((c + sums) % 9) == 0

# Function to find the largest number
def maxNumber(s1, s2, i):
    if s1[0][s1[1]] > s2[s1[1]]:
        return s1
    return [s2, i]

# Function to find the largest integer
# which is divisible by 9 after
# inserting any digit in the N
def findLargestNumber(N, length):
  
    # NOTE: we are using length as the variable name
    # instead of len because len is a predefined method in Python3
    # that we will be using in this program
    # this is a good practice in code

    # Stores the largest integer which is
    # divisible by 9

    maxi = ["", 0]
    # Stores the sum of digits of N
    sums = 0
    for i in range(length):
        # Update the value of sum
        sums += ord(N[i]) - ord("0")

    for i in range(length + 1):
        for ch in range(10):
            # Skip leading zeroes
            if i == 0 and ch == 0:
                continue
            # Check number is divisible by 9
            if isDivBy9(ch, sums):
                # Check maxi is not set
                if len(maxi[0]) == 0:
                    # Set value of maxi
                    maxi = [N[0:i] + str(ch) + N[i::], i]
                else:
                    # Update value of maxi
                    maxi = maxNumber(maxi, N[0:i] + str(ch) + N[i:], i)
    # Print the value of the first
    # element of maxi
    return maxi[0]

# Driver Code
N = "12"
length = len(N)

# Function call
print(findLargestNumber(N, length))

# This code is contributed by phasing17

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

  // Function to check number is divisible by 9
  static bool isDivBy9(char c, int sum)
  {
    return ((((int)c - (int)('0')) + sum) % 9 == 0);
  }

  // Function to find the largest number
  static pair maxNumber(pair s1, String s2, int i)
  {
    if(s1.second == null){
      return s1;
    }
    int x = int.Parse(s1.second);
    if(s1.first == null){
      return s1;
    }

    if (((int)(s1.first[x]) - (int)('0')) > ((int)s2[x] - (int)('0'))){
      return s1;
    }
    pair map = new pair(s2, i.ToString());
    return map;
  }

  // Function to find the largest integer
  // which is divisible by 9 after
  // inserting any digit in the N
  static String findLargestNumber(String N, int len)
  {

    // Stores the largest integer which is
    // divisible by 9
    pair maxi = new pair("", "0");

    // Stores the sum of digits of N
    int sum = 0;
    for (int i = 0 ; i < len ; i++) {

      // Update the value of sum
      sum += ((int)N[i] - (int)('0'));
    }

    for (int i = 0 ; i <= len ; i++) {
      for (char ch = '0' ; ch <= '9' ; ch++) {

        // Skip leading zeroes
        if ((i == 0) && (ch == '0'))
          continue;

        // Check number is divisible by 9
        if (isDivBy9(ch, sum)) {

          // Check maxi is not set
          if ((maxi.first) == "") {

            // Set value of maxi
            maxi.first = N.Substring(0, i) + ch + N.Substring(i);
            maxi.second = i.ToString();
          }
          else {

            // Update value of maxi
            maxi = maxNumber(maxi, N.Substring(0, i) + ch + N.Substring(i), i);
          }
        }
      }
    }

    // Print the value of maxi.first
    return maxi.first;
  }

  public static void Main(string[] args){

    String N = "12";
    int len = N.Length;

    // Function call
    Console.Write(findLargestNumber(N, len));

  }
}

public class pair{
  public String first;
  public String  second;
  public pair(String first, String second){
    this.first = first;
    this.second = second;
  }
}

// This code is contributed by entertain2022.

JavaScript

    <script>
        // JavaScript program for the above approach

        // Function to check number is divisible by 9
        const isDivBy9 = (c, sum) => {
            return ((c + sum) % 9 == 0);
        }

        // Function to find the largest number
        const maxNumber = (s1, s2, i) => {
            if (s1[0][s1[1]] > s2[s1[1]])
                return s1;
            return [s2, i];
        }

        // Function to find the largest integer
        // which is divisible by 9 after
        // inserting any digit in the N
        const findLargestNumber = (N, len) => {
            // Stores the largest integer which is
            // divisible by 9
            let maxi = ["", 0];

            // Stores the sum of digits of N
            let sum = 0;
            for (let i = 0; i < len; i++) {

                // Update the value of sum
                sum += (N.charCodeAt(i) - '0'.charCodeAt(0));
            }

            for (let i = 0; i <= len; i++) {
                for (let ch = 0; ch <= 9; ch++) {

                    // Skip leading zeroes
                    if (i == 0 && ch == 0)
                        continue;

                    // Check number is divisible by 9
                    if (isDivBy9(ch, sum)) {

                        // Check maxi is not set
                        if (maxi[0].length == 0) {

                            // Set value of maxi
                            maxi = [N.substring(0, i) + ch.toString() + N.substring(i), i];
                        }
                        else {

                            // Update value of maxi
                            maxi = maxNumber(maxi,
                                N.substr(0, i) + ch.toString() + N.substr(i), i);
                        }
                    }
                }
            }

            // Print the value of maxi.first
            return maxi[0];
        }

        // Driver Code

        let N = "12";
        let len = N.length;

        // Function call
        document.write(findLargestNumber(N, len));

    // This code is contributed by rakeshsahni

    </script>

Output

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

Find a N-digit number such that it is not divisible by any of its digits

dharanendralv23

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Largest integer divisible by 9 after inserting any digit in N

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