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

Difficulty Level : Hard
Last Updated : 05 Aug, 2021

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:

Either select only a single digit, add a dot and move onto selecting other blocks (further function calls).
Or select two digits at the same time, add a dot and move further.
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 poosible 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 poosible 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 poosible 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";
    }
}

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 poosible 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 poosible 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 poosible 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

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)

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