Convert given string to a valid mobile number

Given a string M consisting of letters, digits and symbols, the task is to convert the string to a valid mobile number by removing all characters except the digits in the following format:

  • Form a substring of 3 digits while the length of the remaining string is greater than 3.
  • Enclose each substring with brackets “()” and separate them with “-“.

If a valid mobile number cannot be obtained, i.e. if the string does not consist of 10 digits, then print -1. Otherwise, print the string obtained.

Examples:

Input: M = “91 234rt5%34*0 3”
Output: “(912)-(345)-(340)-(3)”
Explanation: After removing all extra characters, M = “9123453403”. Therefore, the final string obtained is “(912)-(345)-(340)-(3)”.

Input: M=”9 9 ry7%64 9 7″
Output: “Invalid”
Explanation: After removing extra characters M=”9976497″. Since the length of the string is not equal to 10, the required output is -1.

Approach: Follow the below steps to solve the problem:

  • Initialize a string, say S and append all digits of M in S in the given order.
  • Now, if the length of S is not 10, print “Invalid”, and end the program.
  • Otherwise, if the length of the string S is 10, make a group of 3 characters and enclose it within the brackets “()” and make separate it with “-“.
  • Print final string as S.

Below is the solution for the above problem.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the valid
// and formatted phone number
void Validate(string M)
{
    // Length of given string
    int len = M.size();
 
    // Store digits in temp
    string temp = "";
 
    // Iterate given M
    for (int i = 0; i < len; i++) {
        // If any digit, append it to temp
        if (isdigit(M[i]))
            temp += M[i];
    }
 
    // Find new length of string
    int nwlen = temp.size();
 
    // If length is not equal to 10
    if (nwlen != 10) {
        cout << "Invalid\n";
        return;
    }
 
    // Store final result
    string res = "";
 
    // Make groups of 3 digits and
    // enclose them within () and
    // separate them with "-"
    // 0 to 2 index 1st group
    string x = temp.substr(0, 3);
    res += "(" + x + ")-";
 
    // 3 to 5 index 2nd group
    x = temp.substr(3, 3);
    res += "(" + x + ")-";
 
    // 6 to 8 index 3rd group
    x = temp.substr(6, 3);
    res += "(" + x + ")-";
 
    // 9 to 9 index last group
    x = temp.substr(9, 1);
    res += "(" + x + ")";
 
    // Print final result
    cout << res << "\n";
}
 
// Driver Code
int main()
{
 
    // Given string
    string M = "91 234rt5%34*0 3";
 
    // Function Call
    Validate(M);
}
 
// contributed by ajaykr00kj


Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to print the valid
// and formatted phone number
static void Validate(String M)
{
   
    // Length of given String
    int len = M.length();
 
    // Store digits in temp
    String temp = "";
 
    // Iterate given M
    for (int i = 0; i < len; i++)
    {
       
        // If any digit, append it to temp
        if (Character.isDigit(M.charAt(i)))
            temp += M.charAt(i);
    }
 
    // Find new length of String
    int nwlen = temp.length();
 
    // If length is not equal to 10
    if (nwlen != 10)
    {
        System.out.print("Invalid\n");
        return;
    }
 
    // Store final result
    String res = "";
 
    // Make groups of 3 digits and
    // enclose them within () and
    // separate them with "-"
    // 0 to 2 index 1st group
    String x = temp.substring(0, 3);
    res += "(" + x + ")-";
 
    // 3 to 5 index 2nd group
    x = temp.substring(3, 6);
    res += "(" + x + ")-";
 
    // 6 to 8 index 3rd group
    x = temp.substring(6, 9);
    res += "(" + x + ")-";
 
    // 9 to 9 index last group
    x = temp.substring(9, 10);
    res += "(" + x + ")";
 
    // Print final result
    System.out.print(res+ "\n");
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given String
    String M = "91 234rt5%34*0 3";
 
    // Function Call
    Validate(M);
}
}
 
// This code is contributed by shikhasingrajput


Python3




# Python3 program for the above approach
 
# Function to print valid
# and formatted phone number
def Validate(M):
     
    # Length of given
    lenn = len(M)
 
    # Store digits in temp
    temp = ""
 
    # Iterate given M
    for i in range(lenn):
         
        # If any digit:append it to temp
        if (M[i].isdigit()):
            temp += M[i]
 
    # Find new length of
    nwlenn = len(temp)
 
    # If length is not equal to 10
    if (nwlenn != 10):
        print ("Invalid")
        return
 
    # Store final result
    res = ""
 
    # Make groups of 3 digits and
    # enclose them within () and
    # separate them with "-"
    # 0 to 2 index 1st group
    x = temp[0:3]
    res += "(" + x + ")-"
 
    # 3 to 5 index 2nd group
    x = temp[3 : 3 + 3]
    res += "(" + x + ")-"
 
    # 6 to 8 index 3rd group
    x = temp[6 : 3 + 6]
    res += "(" + x + ")-"
 
    # 9 to 9 index last group
    x = temp[9 : 1 + 9]
    res += "(" + x + ")"
 
    # Print final result
    print(res)
 
# Driver Code
if __name__ == '__main__':
     
    # Given
    M = "91 234rt5%34*0 3"
 
    # Function Call
    Validate(M)
 
# This code is contributed by mohit kumar 29


C#




// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to print the valid
  // and formatted phone number
  static void Validate(string M)
  {
 
    // Length of given String
    int len = M.Length;
 
    // Store digits in temp
    string temp = "";
 
    // Iterate given M
    for (int i = 0; i < len; i++)
    {
 
      // If any digit, append it to temp
      if (Char.IsDigit(M[i]))
        temp += M[i];
    }
 
    // Find new length of String
    int nwlen = temp.Length;
 
    // If length is not equal to 10
    if (nwlen != 10)
    {
      Console.Write("Invalid\n");
      return;
    }
 
    // Store final result
    string res = "";
 
    // Make groups of 3 digits and
    // enclose them within () and
    // separate them with "-"
    // 0 to 2 index 1st group
    string x = temp.Substring(0, 3);
    res += "(" + x + ")-";
 
    // 3 to 5 index 2nd group
    x = temp.Substring(3, 3);
    res += "(" + x + ")-";
 
    // 6 to 8 index 3rd group
    x = temp.Substring(6, 3);
    res += "(" + x + ")-";
 
    // 9 to 9 index last group
    x = temp.Substring(9, 1);
    res += "(" + x + ")";
 
    // Print final result
    Console.WriteLine(res);
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given String
    string M = "91 234rt5%34*0 3";
 
    // Function Call
    Validate(M);
  }
}
 
// This code is contributed by AnkThon


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to print the valid
// and formatted phone number
function Validate(M)
{
    // Length of given String
    let len = M.length;
  
    // Store digits in temp
    let temp = "";
  
    // Iterate given M
    for (let i = 0; i < len; i++)
    {
        
        // If any digit, append it to temp
        if (! isNaN( parseInt(M[i]) ))
            temp += M[i];
    }
  
    // Find new length of String
    let nwlen = temp.length;
  
    // If length is not equal to 10
    if (nwlen != 10)
    {
        document.write("Invalid<br>");
        return;
    }
  
    // Store final result
    let res = "";
  
    // Make groups of 3 digits and
    // enclose them within () and
    // separate them with "-"
    // 0 to 2 index 1st group
    let x = temp.substring(0, 3);
    res += "(" + x + ")-";
  
    // 3 to 5 index 2nd group
    x = temp.substring(3, 6);
    res += "(" + x + ")-";
  
    // 6 to 8 index 3rd group
    x = temp.substring(6, 9);
    res += "(" + x + ")-";
  
    // 9 to 9 index last group
    x = temp.substring(9, 10);
    res += "(" + x + ")";
  
    // Print final result
    document.write(res+ "<br>");
}
 
// Driver Code
// Given String
let M = "91 234rt5%34*0 3";
// Function Call
Validate(M);
 
 
// This code is contributed by patel2127
 
</script>


Output: 

(912)-(345)-(340)-(3)

 

Time Complexity: O(|M|+|S|)
Auxiliary Space: O(|M|+|S|)

 

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!

Related Articles

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button