Program to generate all possible valid IP addresses from given string

30 Jun, 2022
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: 25525511135
Output: [“255.255.11.135”, “255.255.111.35”]
Explanation:
These are the only valid possible
IP addresses.

Input: "25505011535"
Output: []
Explanation: 
We cannot generate a valid IP
address with this string.
First, we will place 3 dots in the given string and then try out all the possible combinations for the 3 dots. Corner case for validity:
For string "25011255255"
25.011.255.255 is not valid as 011 is not valid.
25.11.255.255 is not valid either as you are not
allowed to change the string.
250.11.255.255 is valid.
Approach: Split the string with ‘ . ‘ and then check for all corner cases. Before entering the loop, check the size of the string. Generate all the possible combinations using looping through the string. If IP is found to be valid then return the IP address, else simply return the empty list.Below is the implementation of the above approach:
C++
Java
Python3
Javascript
// C++ program to generate all possible// valid IP addresses from given string#include <bits/stdc++.h>using namespace std; // Function checks whether IP digits// are valid or not.int is_valid(string ip){    // Splitting by "."    vector<string> ips;    string ex = "";    for (int i = 0; i < ip.size(); i++) {        if (ip[i] == '.') {            ips.push_back(ex);            ex = "";        }        else {            ex = ex + ip[i];        }    }    ips.push_back(ex);     // Checking for the corner cases    // cout << ip << endl;    for (int i = 0; i < ips.size(); i++) {        // cout << ips[i] <<endl;        if (ips[i].length() > 3            || stoi(ips[i]) < 0            || stoi(ips[i]) > 255)            return 0;         if (ips[i].length() > 1            && stoi(ips[i]) == 0)            return 0;         if (ips[i].length() > 1            && stoi(ips[i]) != 0            && ips[i][0] == '0')            return 0;    }    return 1;} // Function converts string to IP addressvoid convert(string ip){    int l = ip.length();     // Check for string size    if (l > 12 || l < 4) {        cout << "Not Valid IP Address";    }     string check = ip;    vector<string> ans;     // Generating different combinations.    for (int i = 1; i < l - 2; i++) {        for (int j = i + 1; j < l - 1; j++) {            for (int k = j + 1; k < l; k++) {                check = check.substr(0, k) + "."                        + check.substr(k);                check                    = check.substr(0, j) + "."                      + check.substr(j);                check                    = check.substr(0, i) + "."                      + check.substr(i);                 // cout<< check <<endl;                // Check for the validity of combination                if (is_valid(check)) {                    ans.push_back(check);                    std::cout << check << '\n';                }                check = ip;            }        }    }} // Driver codeint main(){    string A = "25525511135";    string B = "25505011535";     convert(A);    convert(B);     return 0;} // This code is contributed by Harshit
// Java Program to generate all possible// valid IP addresses from given stringimport java.util.*; class GFG {     // Function to restore Ip Addresses    public static ArrayList<String>    restoreIpAddresses(String A)    {        if (A.length() < 3 || A.length() > 12)            return new ArrayList<>();        return convert(A);    }     private static ArrayList<String>    convert(String s)    {        ArrayList<String> l = new ArrayList<>();        int size = s.length();         String snew = s;         for (int i = 1; i < size - 2;             i++) {            for (int j = i + 1;                 j < size - 1; j++) {                for (int k = j + 1;                     k < size; k++) {                    snew                        = snew.substring(0, k) + "."                          + snew.substring(k);                    snew                        = snew.substring(0, j) + "."                          + snew.substring(j);                    snew                        = snew.substring(0, i) + "."                          + snew.substring(i);                     if (isValid(snew)) {                        l.add(snew);                    }                    snew = s;                }            }        }         Collections.sort(l, new Comparator<String>() {            public int compare(String o1, String o2)            {                String a1[] = o1.split("[.]");                String a2[] = o2.split("[.]");                 int result = -1;                for (int i = 0; i < 4                                && result != 0;                     i++) {                    result = a1[i].compareTo(a2[i]);                }                return result;            }        });        return l;    }     private static boolean isValid(String ip)    {        String a[] = ip.split("[.]");        for (String s : a) {            int i = Integer.parseInt(s);            if (s.length() > 3 || i < 0 || i > 255) {                return false;            }            if (s.length() > 1 && i == 0)                return false;            if (s.length() > 1 && i != 0                && s.charAt(0) == '0')                return false;        }         return true;    }     // Driver Code    public static void main(String[] args)    {        System.out.println(            restoreIpAddresses(                "25525511135")                .toString());    }} // This code is contributed by Nidhi Hooda.
# Python3 code to check valid possible IP # Function checks whether IP digits# are valid or not.def is_valid(ip):     # Splitting by "."    ip = ip.split(".")         # Checking for the corner cases    for i in ip:        if (len(i) > 3 or int(i) < 0 or                          int(i) > 255):            return False        if len(i) > 1 and int(i) == 0:            return False        if (len(i) > 1 and int(i) != 0 and            i[0] == '0'):            return False                 return True # Function converts string to IP addressdef convert(s):         sz = len(s)     # Check for string size    if sz > 12:        return []    snew = s    l = []     # Generating different combinations.    for i in range(1, sz - 2):        for j in range(i + 1, sz - 1):            for k in range(j + 1, sz):                snew = snew[:k] + "." + snew[k:]                snew = snew[:j] + "." + snew[j:]                snew = snew[:i] + "." + snew[i:]                                 # Check for the validity of combination                if is_valid(snew):                    l.append(snew)                                     snew = s                     return l # Driver code        A = "25525511135"B = "25505011535" print(convert(A))print(convert(B))
<script> // JavaScript program to generate all possible// valid IP addresses from given string // Function checks whether IP digits// are valid or not.function is_valid(ip){    // Splitting by "."    let ips = new Array();    let ex = "";    for (let i = 0; i < ip.length; i++) {        if (ip[i] == '.') {            ips.push(ex);            ex = "";        }        else {            ex = ex + ip[i];        }    }    ips.push(ex);     // Checking for the corner cases      for(let i = 0; i < ips.length; i++) {                 if (ips[i].length > 3            || parseInt(ips[i]) < 0            || parseInt(ips[i]) > 255)            return 0;         if (ips[i].length > 1            && parseInt(ips[i]) == 0)            return 0;         if (ips[i].length > 1            && parseInt(ips[i]) != 0            && ips[i][0] == '0')            return 0;    }    return 1;} // Function converts string to IP addressfunction convert(ip){    let l = ip.length;     // Check for string size    if (l > 12 || l < 4) {        document.write("Not Valid IP Address");    }     let check = ip;    let ans = new Array();     // Generating different combinations.    for (let i = 1; i < l - 2; i++) {        for (let j = i + 1; j < l - 1; j++) {            for (let k = j + 1; k < l; k++) {                check = check.substring(0, k) + "."                        + check.substring(k,check.length);                check                    = check.substring(0, j) + "."                      + check.substring(j,check.length);                check                    = check.substring(0, i) + "."                      + check.substring(i,check.length);                 // Check for the validity of combination                if (is_valid(check)) {                    ans.push(check);                    document.write(check,"</br>");                }                check = ip;            }        }    }} // Driver code     let A = "25525511135";    let B = "25505011535";     convert(A);    convert(B);     // This code is contributed by shinjanpatra  </script>
255.255.11.135
255.255.111.35
Complexity Analysis:
Time Complexity: O(n^3), where n is the length of the string Three nested traversal of the string is needed, where n is always less than 12.
Auxiliary Space: O(n). As extra space is needed.
Another Efficient Approach (Dynamic Programming): There is a dp approach exist for this problem and can be solved in time complexity O(n*4*3)=O(12n)=O(n) and space complexity O(4n). 
Approach: We know that there are only 4 parts of the IP. We start iterating from the end of string and goes to the start of string. We create a dp 2D-array of size (4 x N). There can be only 2 values in the dp array i.e. 1(true) or 0(false). dp[0][i] tells if we can create 1 part of IP from the substring starting from i to end of string. Similarly, dp[1][i] tells if we can create 2 parts of IP from the substring starting from i to end of string.
After creating the dp array, we start creating the valid IP addresses. We start from the bottom left corner of the 2D dp array. We only iterate 12 times(worst case) but those also will be the valid IP addresses because we only form valid IP addresses. 
Java
C#
// Java Program to generate all possible// valid IP addresses from given stringimport java.util.*; class GFG{    public static ArrayList<String> list;         // Function to restore Ip Addresses    public static ArrayList<String>    restoreIpAddresses(String s)    {        int n = s.length();        list = new ArrayList<>();        if (n < 4 || n > 12)            return list;                // initialize the dp array        int dp[][] = new int[4][n];        for (int i = 0; i < 4; i++)        {            for (int j = n - 1; j >= 0; j--)            {                if (i == 0)                {                    // take the substring                    String sub = s.substring(j);                    if (isValid(sub))                    {                        dp[i][j] = 1;                    }                    else if (j < n - 3)                    {                        break;                    }                }                else                {                    if (j <= n - i)                    {                        for (int k = 1;                             k <= 3 && j + k <= n;                             k++)                        {                            String temp                                = s.substring(j, j + k);                            if (isValid(temp))                            {                                if (j + k < n                                    && dp[i - 1][j + k]                                           == 1)                                {                                    dp[i][j] = 1;                                    break;                                }                            }                        }                    }                }            }        }                 if (dp[3][0] == 0)            return list;                      // Call function createIpfromDp        createIpFromDp(dp, 3, 0, s, "");        return list;    }       public static void createIpFromDp(int dp[][],                                      int r,                                      int c, String s,                                      String ans)    {        if (r == 0)        {            ans += s.substring(c);            list.add(ans);            return;        }        for (int i = 1;             i <= 3 && c + i < s.length();             i++)        {            if (isValid(s.substring(c, c + i))                && dp[r - 1] == 1)            {                createIpFromDp(dp, r - 1, c + i, s,                               ans +                               s.substring(c, c + i)                               + ".");            }        }    }          private static boolean isValid(String ip)    {        String a[] = ip.split("[.]");        for (String s : a)        {            int i = Integer.parseInt(s);            if (s.length() > 3 || i < 0 || i > 255)            {                return false;            }            if (s.length() > 1 && i == 0)                return false;            if (s.length() > 1 && i != 0                && s.charAt(0) == '0')                return false;        }         return true;    }     // Driver Code    public static void main(String[] args)    {        // Function call        System.out.println(            restoreIpAddresses("25525511135").toString());    }} // This code is contributed by Nidhi Hooda.
// Include namespace systemusing System;using System.Collections.Generic; public class GFG {  static List<String> list;   // Function to restore Ip Addresses  public static List<String> restoreIpAddresses(String s)  {    var n = s.Length;    list = new List<String>();    if (n < 4 || n > 12) {      return list;    }         // initialize the dp array    int[, ] dp = new int[4, n];    for (int i = 0; i < 4; i++) {      for (int j = n - 1; j >= 0; j--) {        if (i == 0)        {                     // take the substring          var sub = s.Substring(j);          if (GFG.isValid(sub)) {            dp[i, j] = 1;          }          else if (j < n - 3) {            break;          }        }        else {          if (j <= n - i) {            for (int k = 1;                 k <= 3 && j + k <= n; k++) {              var temp                = s.Substring(j, j + k - j);              if (GFG.isValid(temp)) {                if (j + k < n                    && dp[i - 1, j + k]                    == 1) {                  dp[i, j] = 1;                  break;                }              }            }          }        }      }    }    if (dp[3, 0] == 0) {      return list;    }         // Call function createIpfromDp    GFG.createIpFromDp(dp, 3, 0, s, "");    return list;  }  public static void createIpFromDp(int[, ] dp, int r,                                    int c, String s,                                    String ans)  {    if (r == 0) {      ans += s.Substring(c);      list.Add(ans);      return;    }    for (int i = 1; i <= 3 && c + i < s.Length; i++) {      if (GFG.isValid(s.Substring(c, c + i - c))          && dp[r - 1, c + i] == 1) {        GFG.createIpFromDp(          dp, r - 1, c + i, s,          ans + s.Substring(c, c + i - c) + ".");      }    }  }  private static bool isValid(String ip)  {    String[] a = ip.Split("[.]");    foreach(String s in a)    {      var i = int.Parse(s);      if (s.Length > 3 || i < 0 || i > 255) {        return false;      }      if (s.Length > 1 && i == 0) {        return false;      }      if (s.Length > 1 && i != 0 && s[0] == '0') {        return false;      }    }    return true;  }   // Driver Code  public static void Main(String[] args)  {     // Function call    Console.WriteLine(string.Join(      ",", (GFG.restoreIpAddresses("25525511135"))      .ToArray()));  }} // This code is contributed by Aarti_Rathi
[255.255.11.135, 255.255.111.35]
Complexity Analysis:
Time Complexity: O(n), where n is the length of the string. The dp array creation would take O(4*n*3) = O(12n) = O(n). Valid IP creation from dp array would take O(n).
Auxiliary Space: O(n). As dp array has extra space of size (4 x n). It means O(4n).
Another Approach: (Using Recursion)
C++
Java
Python3
Javascript
#include <iostream>#include <vector>using namespace std; void solve(string s, int i, int j, int level, string temp,           vector<string>& res){    if (i == (j + 1) && level == 5) {        res.push_back(temp.substr(1));    }     // Digits of a number ranging 0-255 can lie only between    // 0-3    for (int k = i; k < i + 3 && k <= j; k++) {        string ad = s.substr(i, k - i + 1);         // Return if string starting with '0' or it is        // greater than 255.        if ((s[i] == '0'&&ad.size()>1) || stol(ad) > 255)            return;         // Recursively call for another level.        solve(s, k + 1, j, level + 1, temp + '.' + ad, res);    }} int main(){    string s = "25525511135";    int n = s.length();     vector<string> ans;     solve(s, 0, n - 1, 1, "", ans);     for (string s : ans)        cout << s << endl;     return 0;}
/*package whatever //do not write package name here */ import java.io.*;import java.util.*; class GFG {    static void solve(String s, int i, int j, int level, String temp,           ArrayList<String>res){    if (i == (j + 1) && level == 5) {        res.add(temp.substring(1));    }     // Digits of a number ranging 0-255 can lie only between    // 0-3    for (int k = i; k < i + 3 && k <= j; k++) {        String ad = s.substring(i, k + 1);         // Return if string starting with '0' or it is        // greater than 255.        if ((s.charAt(i) == '0'&& ad.length()>1 ) || Integer.valueOf(ad) > 255)            return;         // Recursively call for another level.        solve(s, k + 1, j, level + 1, temp + '.' + ad, res);    }}  /* Driver program to test above function*/public static void main(String args[]){    String s = "25525511135";    int n = s.length();     ArrayList<String> ans = new ArrayList<>();     solve(s, 0, n - 1, 1, "", ans);     for (String s1 : ans)        System.out.println(s1);}} // This code is contributed by shinjanpatra.
# Python code for the above approachdef solve(s, i, j, level, temp, res):     if (i == (j + 1) and level == 5):        res.append(temp[1:])     # Digits of a number ranging 0-255 can lie only between    # 0-3    k = i    while(k < i + 3 and k <= j):                 ad = s[i: k + 1]         # Return if string starting with '0' or it is        # greater than 255.        if ((s[i] == '0' and len(ad) > 1) or int(ad) > 255):            return         # Recursively call for another level.        solve(s, k + 1, j, level + 1, temp + '.' + ad, res)            k += 1 # driver codes = "25525511135"n = len(s)ans = []solve(s, 0, n - 1, 1, "", ans)for s in ans:    print(s) # This code is contributed by shinjanpatra
<script> // Simple JavaScript program function solve(s,i,j,level, temp,res){    if (i == (j + 1) && level == 5) {        res.push(temp.substring(1,));    }     // Digits of a number ranging 0-255 can lie only between    // 0-3    for (let k = i; k < i + 3 && k <= j; k++) {        let ad = s.substring(i, k + 1);         // Return if string starting with '0' or it is        // greater than 255.        if ((s[i] == '0' && ad.length > 1) || parseInt(ad) > 255)            return;         // Recursively call for another level.        solve(s, k + 1, j, level + 1, temp + '.' + ad, res);       }} // driver code let s = "25525511135";let n = s.length; let ans = []; solve(s, 0, n - 1, 1, "", ans); for (let s of ans)    document.write(s,"</br>"); // This code is contributed by shinjanpatra </script>
255.255.11.135
255.255.111.35
 
nidhi hooda
Harshit-Agarwal
shubham_singh
andrew1234
Ripunjoy Medhi
vikas_motiani
reena_yadav
jatin1233
yashparsana135
shinjanpatra
simmytarika5
AshokJaiswal
adi1212
Python list-programs
Python string-programs
python-list
python-string
Python
Strings
python-list
Strings
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.