Program to generate all possible valid IP addresses from given string | Set 2

Given a string containing only digits, restore it by returning all possible valid IP address combinations. 
A valid IP address must be in the form of A.B.C.D, where A, B, C and D are numbers from 0 – 255. The numbers cannot be 0 prefixed unless they are 0.
Examples:

Input: str = “25525511135” 
Output:
255.255.11.135 
255.255.111.35
Input: str = “11111011111” 
Output:
111.110.11.111 
111.110.111.11 

Approach: This problem can be solved using backtracking. In each call we have three options to create a single block of numbers of a valid ip address: 

1. Either select only a single digit, add a dot and move onto selecting other blocks (further function calls).
2. Or select two digits at the same time, add a dot and move further.
3. Or select three consecutive digits and move for the next block.

At the end of the fourth block, if all the digits have been used and the address generated is a valid ip-address then add it to the results and then backtrack by removing the digits selected in the previous call.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
#include <vector>
using namespace std;

// Function to get all the valid ip-addresses
void GetAllValidIpAddress(vector<string>& result,
                          string givenString, int index,
                          int count, string ipAddress)
{

    // If index greater than givenString size
    // and we have four block
    if (givenString.size() == index && count == 4) {

        // Remove the last dot
        ipAddress.pop_back();

        // Add ip-address to the results
        result.push_back(ipAddress);
        return;
    }

    // To add one index to ip-address
    if (givenString.size() < index + 1)
        return;

    // Select one digit and call the
    // same function for other blocks
    ipAddress = ipAddress
                + givenString.substr(index, 1) + '.';
    GetAllValidIpAddress(result, givenString, index + 1,
                         count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove two index (one for the digit
    // and other for the dot) from the end
    ipAddress.erase(ipAddress.end() - 2, ipAddress.end());

    // Select two consecutive digits and call
    // the same function for other blocks
    if (givenString.size() < index + 2
        || givenString[index] == '0')
        return;
    ipAddress = ipAddress + givenString.substr(index, 2) + '.';
    GetAllValidIpAddress(result, givenString, index + 2,
                         count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove three index from the end
    ipAddress.erase(ipAddress.end() - 3, ipAddress.end());

    // Select three consecutive digits and call
    // the same function for other blocks
    if (givenString.size() < index + 3
        || stoi(givenString.substr(index, 3)) > 255)
        return;
    ipAddress += givenString.substr(index, 3) + '.';
    GetAllValidIpAddress(result, givenString, index + 3,
                         count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove four index from the end
    ipAddress.erase(ipAddress.end() - 4, ipAddress.end());
}

// Driver code
int main()
{
    string givenString = "25525511135";

    // Fill result vector with all valid ip-addresses
    vector<string> result;
    GetAllValidIpAddress(result, givenString, 0, 0, "");

    // Print all the generated ip-addresses
    for (int i = 0; i < result.size(); i++) {
        cout << result[i] << "\n";
    }
}

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;

class GFG
{

  // Function to get all the valid ip-addresses
  static void GetAllValidIpAddress(ArrayList<String>result,
                                   String givenString,int index,
                                   int count, String ipAddress){

    // If index greater than givenString size
    // and we have four block
    if (givenString.length() == index && count == 4){

      // Remove the last dot
      ipAddress =  ipAddress.substring(0,ipAddress.length()-1);

      // Add ip-address to the results
      result.add(ipAddress);
      return;
    }

    // To add one index to ip-address
    if (givenString.length() < index + 1)
      return;

    // Select one digit and call the
    // same function for other blocks
    ipAddress = (ipAddress + givenString.substring(index , index + 1) + '.');

    GetAllValidIpAddress(result, givenString, index + 1, count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove two index (one for the digit
    // and other for the dot) from the end
    ipAddress = ipAddress.substring(0,ipAddress.length() - 2);

    // Select two consecutive digits and call
    // the same function for other blocks
    if (givenString.length() < index + 2 ||
        givenString.charAt(index) == '0')
      return;

    ipAddress = ipAddress + givenString.substring(index,index + 2) + '.';
    GetAllValidIpAddress(result, givenString, index + 2,
                         count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove three index from the end
    ipAddress = ipAddress.substring(0,ipAddress.length() - 3);

    // Select three consecutive digits and call
    // the same function for other blocks
    if (givenString.length() < index + 3 ||
        Integer.valueOf(givenString.substring(index,index + 3)) > 255)
      return;
    ipAddress += givenString.substring(index,index + 3) + '.';
    GetAllValidIpAddress(result, givenString,
                         index + 3, count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove four index from the end
    ipAddress = ipAddress.substring(0,ipAddress.length()-4);
  }
  public static void main (String[] args) {
    String givenString = "25525511135";

    // Fill result vector with all valid ip-addresses
    ArrayList<String>result = new ArrayList<String>() ;
    String ipAddress = "";
    GetAllValidIpAddress(result, givenString, 0, 0,ipAddress);

    // Print all the generated ip-addresses
    for(int i = 0; i < result.size(); i++)
      System.out.println(result.get(i));
  }
}

// This code is contributed by shinjanpatra.

Python3

# Python3 implementation of the approach

# Function to get all the valid ip-addresses
def GetAllValidIpAddress(result, givenString,
                         index, count, ipAddress) :

    # If index greater than givenString size
    # and we have four block
    if (len(givenString) == index and count == 4) :

        # Remove the last dot
        ipAddress.pop();

        # Add ip-address to the results
        result.append(ipAddress);
        return;

    # To add one index to ip-address
    if (len(givenString) < index + 1) :
        return;

    # Select one digit and call the
    # same function for other blocks
    ipAddress = (ipAddress +
                 givenString[index : index + 1] + ['.']);

    GetAllValidIpAddress(result, givenString, index + 1,
                                 count + 1, ipAddress);

    # Backtrack to generate another possible ip address
    # So we remove two index (one for the digit
    # and other for the dot) from the end
    ipAddress = ipAddress[:-2];

    # Select two consecutive digits and call
    # the same function for other blocks
    if (len(givenString) < index + 2 or
            givenString[index] == '0') :
        return;

    ipAddress = ipAddress + givenString[index:index + 2] + ['.'];
    GetAllValidIpAddress(result, givenString, index + 2,
                                  count + 1, ipAddress);

    # Backtrack to generate another possible ip address
    # So we remove three index from the end
    ipAddress = ipAddress[:-3];

    # Select three consecutive digits and call
    # the same function for other blocks
    if (len(givenString)< index + 3 or
        int("".join(givenString[index:index + 3])) > 255) :
        return;
    ipAddress += givenString[index:index + 3] + ['.'];
    GetAllValidIpAddress(result, givenString,
                         index + 3, count + 1, ipAddress);

    # Backtrack to generate another possible ip address
    # So we remove four index from the end
    ipAddress = ipAddress[:-4];

# Driver code
if __name__ == "__main__" :
    givenString = list("25525511135");

    # Fill result vector with all valid ip-addresses
    result = [] ;
    GetAllValidIpAddress(result, givenString, 0, 0, []);

    # Print all the generated ip-addresses
    for i in range(len(result)) :
        print("".join(result[i]));

# This code is contributed by Ankitrai01

C#

// C# implementation of the above approach.
using System;
using System.Collections.Generic;

public class GFG {

    // Function to get all the valid ip-addresses
    static void GetAllValidIpAddress(List<string> result,
                                     string givenString,
                                     int index, int count,
                                     string ipAddress)
    {

        // If index greater than givenString size
        // and we have four block
        if (givenString.Length == index && count == 4) {

            // Remove the last dot
            ipAddress
                = ipAddress.Remove(ipAddress.Length - 1, 1);

            // Add ip-address to the results
            result.Add(ipAddress);

            return;
        }

        // To add one index to ip-address
        if (givenString.Length < index + 1) {
            return;
        }

        // Select one digit and call the
        // same function for other blocks
        ipAddress = ipAddress
                    + givenString.Substring(index, 1) + '.';
        GetAllValidIpAddress(result, givenString, index + 1,
                             count + 1, ipAddress);

        // Backtrack to generate another possible ip address
        // So we remove two index (one for the digit
        // and other for the dot) from the end
        ipAddress
            = ipAddress.Remove(ipAddress.Length - 2, 2);

        // Select two consecutive digits and call
        // the same function for other blocks
        if (givenString.Length < index + 2
            || givenString[index] == '0') {
            return;
        }

        ipAddress = ipAddress
                    + givenString.Substring(index, 2) + '.';
        GetAllValidIpAddress(result, givenString, index + 2,
                             count + 1, ipAddress);

        // Backtrack to generate another possible ip address
        // So we remove three index from the end
        ipAddress
            = ipAddress.Remove(ipAddress.Length - 3, 3);

        // Select three consecutive digits and call
        // the same function for other blocks
        if (givenString.Length < index + 3
            || Int64.Parse(givenString.Substring(index, 3))
                   > 255) {
            return;
        }

        ipAddress = ipAddress
                    + givenString.Substring(index, 3) + '.';

        GetAllValidIpAddress(result, givenString, index + 3,
                             count + 1, ipAddress);

        // Backtrack to generate another possible ip address
        // So we remove four index from the end
        ipAddress
            = ipAddress.Remove(ipAddress.Length - 4, 4);
    }

    static public void Main()
    {
        string givenString = "25525511135";

        // Fill result vector with all valid ip-addresses
        List<string> result = new List<string>();
        GetAllValidIpAddress(result, givenString, 0, 0, "");

        // Print all the generated ip-addresses
        for (int i = 0; i < result.Count; i++) {
            System.Console.WriteLine(result[i]);
        }
    }
}

// The  code is contributed by Gautam goel.

Javascript

<script>

// JavaScript implementation of the approach

// Function to get all the valid ip-addresses
function GetAllValidIpAddress(result, givenString, index, count, ipAddress){

    // If index greater than givenString size
    // and we have four block
    if (givenString.length == index && count == 4){

        // Remove the last dot
        ipAddress =  ipAddress.substring(0,ipAddress.length-1);

        // Add ip-address to the results
        result.push(ipAddress);
        return;
    }

    // To add one index to ip-address
    if (givenString.length < index + 1)
        return;

    // Select one digit and call the
    // same function for other blocks
    ipAddress = (ipAddress + givenString.substring(index , index + 1) + '.');

    GetAllValidIpAddress(result, givenString, index + 1, count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove two index (one for the digit
    // and other for the dot) from the end
    ipAddress = ipAddress.substring(0,ipAddress.length - 2);

    // Select two consecutive digits and call
    // the same function for other blocks
    if (givenString.length < index + 2 ||
            givenString[index] == '0')
        return;

    ipAddress = ipAddress + givenString.substring(index,index + 2) + '.';
    GetAllValidIpAddress(result, givenString, index + 2,
                                  count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove three index from the end
    ipAddress = ipAddress.substring(0,ipAddress.length - 3);

    // Select three consecutive digits and call
    // the same function for other blocks
    if (givenString.length < index + 3 ||
        parseInt(givenString.substring(index,index + 3)) > 255)
        return;
    ipAddress += givenString.substring(index,index + 3) + ['.'];
    GetAllValidIpAddress(result, givenString,
                         index + 3, count + 1, ipAddress);

    // Backtrack to generate another possible ip address
    // So we remove four index from the end
    ipAddress = ipAddress.substring(0,ipAddress.length-4);
}

// Driver code

let givenString = "25525511135";

// Fill result vector with all valid ip-addresses
let result = [] ;
GetAllValidIpAddress(result, givenString, 0, 0, []);

// Print all the generated ip-addresses
for(let i=0;i<result.length;i++)
    document.write(result[i],"</br>");

// This code is contributed by shinjanpatra

</script>

Output:

255.255.11.135
255.255.111.35

Time Complexity: O(1), Since the total number of IP addresses are constant
Auxiliary Space: O(1)