CPMagical CP Library

Problem solution · C++

Validate IP Address

Validate IP Address: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Validate IP Address, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 52 lines of C++ from the credited upstream file 468.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 4 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeValidate IP Address · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string validIPAddress(string queryIP) {    string digit;    istringstream iss(queryIP);     if (ranges::count(queryIP, '.') == 3) {      for (int i = 0; i < 4; ++i)  // Make sure that we have four parts.        if (!getline(iss, digit, '.') || !isIPv4(digit))          return "Neither";      return "IPv4";    }     if (ranges::count(queryIP, ':') == 7) {      for (int i = 0; i < 8; ++i)  // Make sure that we have eight parts.        if (!getline(iss, digit, ':') || !isIPv6(digit))          return "Neither";      return "IPv6";    }     return "Neither";  }  private:  static inline string validIPv6Chars = "0123456789abcdefABCDEF";   bool isIPv4(const string& digit) {    if (digit.empty() || digit.length() > 3)      return false;    if (digit.length() > 1 && digit[0] == '0')      return false;     for (const char c : digit)      if (c < '0' || c > '9')        return false;     const int num = stoi(digit);    return 0 <= num && num <= 255;  }   bool isIPv6(const string& digit) {    if (digit.empty() || digit.length() > 4)      return false;     for (const char c : digit)      if (validIPv6Chars.find(c) == string::npos)        return false;     return true;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗