93. Restore IP Addresses LeetCode Solution

In this guide, you will get 93. Restore IP Addresses LeetCode Solution with the best time and space complexity. The solution to Restore IP Addresses problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Restore IP Addresses solution in C++
Restore IP Addresses solution in Java
Restore IP Addresses solution in Python
Additional Resources

93. Restore IP Addresses LeetCode Solution image

Problem Statement of Restore IP Addresses

A 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 “192.168@1.1” 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:

1 <= s.length <= 20
s consists of digits only.

Complexity Analysis

Time Complexity: O(3^4)
Space Complexity: O(1)

93. Restore IP Addresses LeetCode 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-bounds
      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, std::move(path), ans);
      path.pop_back();
    }
  }
};
/* code provided by PROGIEZ */

93. Restore IP Addresses LeetCode 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-bounds
        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);
    }
  }
}
// code provided by PROGIEZ

93. Restore IP Addresses LeetCode 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-bounds
        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
 # code by PROGIEZ