Restore IP Addresses LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Restore IP Addresses in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemRestore IP Addresses– LeetCode Problem

Restore IP Addresses– LeetCode Problem

Problem:

valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.

  • For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245""192.168.1.312" and "[email protected]" are invalid IP addresses.

Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.

Example 1:

Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]

Example 2:

Input: s = "0000"
Output: ["0.0.0.0"]

Example 3:

Input: s = "101023"
Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]

Constraints:

  • 0 <= s.length <= 20
  • s consists of digits only.
Restore IP Addresses– LeetCode Solutions
Restore IP Addresses Solution in C++:
class Solution {
 public:
  vector<string> restoreIpAddresses(const string& s) {
    vector<string> ans;

    dfs(s, 0, {}, ans);

    return ans;
  }

 private:
  void dfs(const string& s, int start, vector<string>&& path,
           vector<string>& ans) {
    if (path.size() == 4 && start == s.length()) {
      ans.push_back(path[0] + "." + path[1] + "." + path[2] + "." + path[3]);
      return;
    }
    if (path.size() == 4 || start == s.length())
      return;

    for (int length = 1; length <= 3; ++length) {
      if (start + length > s.length())
        return;  // out of bound
      if (length > 1 && s[start] == '0')
        return;  // leading '0'
      const string& num = s.substr(start, length);
      if (stoi(num) > 255)
        return;
      path.push_back(num);
      dfs(s, start + length, move(path), ans);
      path.pop_back();
    }
  }
};
Restore IP Addresses Solution in Java:
class Solution {
  public List<String> restoreIpAddresses(final String s) {
    List<String> ans = new ArrayList<>();

    dfs(s, 0, new ArrayList<>(), ans);

    return ans;
  }

  private void dfs(final String s, int start, List<String> path, List<String> ans) {
    if (path.size() == 4 && start == s.length()) {
      ans.add(String.join(".", path));
      return;
    }
    if (path.size() == 4 || start == s.length())
      return;

    for (int length = 1; length <= 3; ++length) {
      if (start + length > s.length()) // out of bound
        return;
      if (length > 1 && s.charAt(start) == '0') // leading '0'
        return;
      final String num = s.substring(start, start + length);
      if (Integer.parseInt(num) > 255)
        return;
      path.add(num);
      dfs(s, start + length, path, ans);
      path.remove(path.size() - 1);
    }
  }
}
Restore IP Addresses Solution in Python:
class Solution:
  def restoreIpAddresses(self, s: str) -> List[str]:
    ans = []

    def dfs(start: int, path: List[int]) -> None:
      if len(path) == 4 and start == len(s):
        ans.append(path[0] + '.' + path[1] + '.' + path[2] + '.' + path[3])
        return
      if len(path) == 4 or start == len(s):
        return

      for length in range(1, 4):
        if start + length > len(s):
          return  # out of bound
        if length > 1 and s[start] == '0':
          return  # leading '0'
        num = s[start: start + length]
        if int(num) > 255:
          return
        dfs(start + length, path + [num])

    dfs(0, [])

    return ans

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