Program to insert dashes between two adjacent odd digits in given Number

Given a large number in form of string N, the task is to insert a dash between two adjacent odd digits in the given number in form of strings.

Examples:

Input: N = 1745389 
Output: 1-745-389 
Explanation:
In string str, str[0] and str[1] both are the odd numbers in consecutive, so insert a dash between them.
Input: N = 34657323128437 
Output: 3465-7-323-12843-7

Bitwise Approach:

1. Traverse the whole string of numbers character by character.
2. Compare every consecutive character using logical Bitwise OR and AND operators.
3. If two consecutive characters of the string are odd, insert a dash (-) in them and check for the next two consecutive characters.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
#include <string>
using namespace std;

// Function to check if char ch is
// odd or not
bool checkOdd(char ch)
{
    return ((ch - '0') & 1);
}

// Function to insert dash - between
// any 2 consecutive digit in string str
string Insert_dash(string num_str)
{

    string result_str = num_str;

    // Traverse the string character
    // by character
    for (int x = 0;
         x < num_str.length() - 1; x++) {

        // Compare every consecutive
        // character with the odd value
        if (checkOdd(num_str[x])
            && checkOdd(num_str[x + 1])) {

            result_str.insert(x + 1, "-");
            num_str = result_str;
            x++;
        }
    }

    // Print the resultant string
    return result_str;
}

// Driver Code
int main()
{

    // Given number in form of string
    string str = "1745389";

    // Function Call
    cout << Insert_dash(str);

    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to check if char ch is
// odd or not
static boolean checkOdd(char ch)
{
    return ((ch - '0') & 1) != 0 ?
            true : false;
}

// Function to insert dash - between
// any 2 consecutive digit in string str
static String Insert_dash(String num_str)
{
    StringBuilder result_str = new StringBuilder(num_str);

    // Traverse the string character
    // by character
    for(int x = 0; x < num_str.length() - 1; x++)
    {

        // Compare every consecutive
        // character with the odd value
        if (checkOdd(num_str.charAt(x)) &&
            checkOdd(num_str.charAt(x + 1)))
        {
            result_str.insert(x + 1, "-");
            num_str = result_str.toString();
            x++;
        }
    }

    // Print the resultant string
    return result_str.toString();
}

// Driver Code
public static void main(String[] args)
{

    // Given number in form of string
    String str = "1745389";

    // Function call
    System.out.println(Insert_dash(str));
}
}

// This code is contributed by rutvik_56

Python3

# Python3 program for the above approach

# Function to check if char ch is
# odd or not
def checkOdd(ch):

    return ((ord(ch) - 48) & 1)

# Function to insert dash - between
# any 2 consecutive digit in string str
def Insert_dash(num_str):

    result_str = num_str

    # Traverse the string character
    # by character
    x = 0
    while(x < len(num_str) - 1):

        # Compare every consecutive
        # character with the odd value
        if (checkOdd(num_str[x]) and
            checkOdd(num_str[x + 1])):

            result_str = (result_str[:x + 1] + '-' +
                          result_str[x + 1:])
            num_str = result_str
            x += 1
        x += 1

    # Print the resultant string
    return result_str

# Driver Code

# Given number in form of string
str = "1745389"

# Function call
print(Insert_dash(str))

# This code is contributed by vishu2908

C#

// C# program to implement
// the above approach
using System;
using System.Text;
class GFG{

// Function to check if char ch is
// odd or not
static bool checkOdd(char ch)
{
    return ((ch - '0') & 1) != 0 ?
            true : false;
}

// Function to insert dash - between
// any 2 consecutive digit in string str
static String Insert_dash(String num_str)
{
    StringBuilder result_str = new StringBuilder(num_str);

    // Traverse the string character
    // by character
    for(int x = 0; x < num_str.Length - 1; x++)
    {

        // Compare every consecutive
        // character with the odd value
        if (checkOdd(num_str[x]) &&
            checkOdd(num_str[x + 1]))
        {
            result_str.Insert(x + 1, "-");
            num_str = result_str.ToString();
            x++;
        }
    }

    // Print the resultant string
    return result_str.ToString();
}

// Driver Code
public static void Main(String[] args)
{

    // Given number in form of string
    String str = "1745389";

    // Function call
    Console.WriteLine(Insert_dash(str));
}
}

// This code is contributed by Rajput-Ji

Javascript

// Javascript program for the above approach

// Function to check if char ch is
// odd or not
function checkOdd(ch)
{
    return (parseInt(ch) & 1);
}

// Function to insert dash - between
// any 2 consecutive digit in string str
function Insert_dash(num_str)
{

    let result_str = "";

    // Traverse the string character
    // by character
    for (let x = 0;
         x < num_str.length - 1; x++) {

        // Compare every consecutive
        // character with the odd value
        if (checkOdd(num_str[x])
            && checkOdd(num_str[x + 1])) {

            result_str+=(num_str[x]+ "-");
        }
        else{
            result_str+=num_str[x];
        }

    }

    result_str+=num_str[num_str.length-1];
    // Print the resultant string
    return result_str;
}

// Driver Code
    // Given number in form of string
    let str = "1745389";

    // Function Call
    document.write(Insert_dash(str));

Output:

1-745-389

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

Regular Expression Approach:

The given problem can be solved using Regular Expression. The RE for this problem will be: 

(?<=[13579])(?=[13579])
The given RE matches between odd numbers. We can replace the matched part of zero width with a dash, i.e.
str = str.replaceAll(“(?<=[13579])(?=[13579])”, “-“);

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <iostream>
#include <regex>
using namespace std;


// Function to insert dash - between
// any 2 consecutive odd digit
string Insert_dash(string str)
{

  // Get the regex to be checked
  const regex pattern("([13579])([13579])");

  // Replaces the matched value
  // (here dash) with given string
  return regex_replace(str, pattern, "$1-$2");;
}

// Driver Code
int main()
{
  string str = "1745389";
  cout << Insert_dash(str);
  return 0;
}

// This code is contributed by yuvraj_chandra

Java

// Java program for the above approach

import java.util.regex.*;

public class GFG {

    // Function to insert dash - between
    // any 2 consecutive odd digit
    public static String Insert_dash(String str)
    {

        // Get the regex to be checked
        String regex = "(?<=[13579])(?=[13579])";

        // Create a pattern from regex
        Pattern pattern = Pattern.compile(regex);

        // Create a matcher for the input String
        Matcher matcher
            = pattern.matcher(str);

        // Get the String to be replaced,
        // i.e. here dash
        String stringToBeReplaced = "-";
        StringBuilder builder
            = new StringBuilder();

        // Replace every matched pattern
        // with the target String
        // using replaceAll() method
        return (matcher
                    .replaceAll(stringToBeReplaced));
    }

    // Driver Code
    public static void main(String[] args)
    {

        // Given number in form of string
        String str = "1745389";

        // Function Call
        System.out.println(Insert_dash(str));
    }
}

Python

# Python program for the above approach
import re

# Function to insert dash - between
# any 2 consecutive odd digit
def Insert_dash(str):

    # Get the regex to be checked
    regex = "(?<=[13579])(?=[13579])"

    return re.sub(regex,'\1-\2', str)

# Driver Code

# Given number in form of string
str = "1745389"

# Function Call
print(Insert_dash(str))

# This code is contributed by yuvraj_chandra

C#

// C# Program to implement
// the above approach
using System;
using System.Text.RegularExpressions;
public class GFG {

  // Function to insert dash - between
  // any 2 consecutive odd digit
  static string Insert_dash(string str)
  {

    // Get the regex to be checked
    string pattern="([13579])([13579])";

    // Replaces the matched value
    // (here dash) with given string
    return Regex.Replace(str, pattern, "$1-$2");;
  }

  // Driver Code
  public static void Main()
  {
    string str = "1745389";
    Console.WriteLine(Insert_dash(str));
  }
}

// This code is contributed by Aman Kumar.

Javascript

// Javascript Program to implement
// the above approach

// Function to insert dash - between
// any 2 consecutive odd digit
function Insert_dash(str)
{

// Get the regex to be checked
let regex = new RegExp("([13579])([13579])",'g');

// Replaces the matched value
// (here dash) with given string
return str.replace(regex, '$1-$2');
}

// Driver Code

let str = "1745389";
console.log(Insert_dash(str));

// This code is contributed by Pushpesh Raj.

Output:

1-745-389

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