Defanging an IP Address Leetcode Solution

Table of Contents

Problem Statement

In this problem, we are given an IP Address. We just have to convert it into a Defanged IP Address i.e. in our output string, all the “.” are converted to “[.]”.

Defanging an IP Address Leetcode Solution

Example

#1:

address = "1.1.1.1"

"1[.]1[.]1[.]1"

#2:

address = "255.100.50.0"

"255[.]100[.]50[.]0"

Approach 1 (Using String Stream/Builder)

For this problem we can use simple string stream or builder class to modify the given string.
We can use a string builder (in case of java) and string stream (in case of C++) to convert given string to output string.
We will traverse the input string from left to right. If any character is ‘.’ then, we will append “[.]” in output string. Otherwise we will simply append the character in output string too.

Algorithm

Create an empty string stream or builder.
Run a for loop to traverse each character of given string.
For each character in string. If character is ‘.” then append “[.]” to string builder. Else append the same character to string builder.
Convert the stream/builder to string and return it.

Implementation for Defanging an IP Address Leetcode Solution

C++ Program

#include <iostream>
#include <sstream>
using namespace std;
string defangIPaddr(string address) 
{
    std::stringstream ss;

    for(int i=0;i<address.length();i++){
        if(address[i]=='.'){
            ss<<"[.]";//replacing . with [.]
        }else{
            ss<<address[i];
        }
    }
    return ss.str();
}

int main()
{
    cout << defangIPaddr("1.1.1.1");
}

1[.]1[.]1[.]1

Java Program

import java.util.*;
import java.lang.*;

class Solution
{  
    public static String defangIPaddr(String address) 
    {
        StringBuilder sb=new StringBuilder();
        for(int i=0;i<address.length();i++){
            if(address.charAt(i)=='.'){
                sb.append("[.]");
            }else{
                sb.append(address.charAt(i));
            }
        }
        return sb.toString();
    }
    
    public static void main(String args[])
    {
        System.out.println(defangIPaddr("1.1.1.1"));
    }
}

1[.]1[.]1[.]1

Complexity Analysis for Defanging an IP Address Leetcode Solution

Time Complexity

O(n) : we are traversing the given string linearly, Thus it’s O(n).

Space Complexity

O(n) : In case of java we have used string builder which is linear extra space and in case of C++ we have used string stream thus space complexity is O(n).

Approach 2 (Using In-built Function)

We will use regex replace function in C++ and replace all function in java:

We can use regex_replace function of C++ to replace all “.” to “[.]”.
Also, in case of java, we can use replaceAll() function to replace all “.” to “[.]”.