Ramiro N. Cosa

Validador de Ipv4 - FreeCodeCamp #25 Daily Challenge

2 min

IPv4 Validator: Analysis and Explanation

Problem Statement

Given a string, determine if it represents a valid IPv4 address. An IPv4 address:

  • Has four integer numbers separated by dots.
  • Each number is between 0 and 255 (inclusive).
  • No leading zeros (“0” is valid, “01” is not).
  • Only contains digits (0-9).

Initial Analysis

Key Test Cases

  • 192.168.1.1 → ✅ Valid
  • 255.255.255.255 → ✅ Valid (upper limit)
  • 0.0.0.0 → ✅ Valid (lower limit)
  • 192.168.01.1 → ❌ Invalid (leading zero)
  • 256.100.100.100 → ❌ Invalid (out of range)
  • 192.168.1 → ❌ Invalid (missing blocks)
  • 192.168.1.1.1 → ❌ Invalid (extra blocks)
  • 192.168.a.1 → ❌ Invalid (non-numeric)
  • 192.168.1.-1 → ❌ Invalid (negative)
  • 192.168.1. 1 → ❌ Invalid (space)
  • 192.168.1.01 → ❌ Invalid (leading zero)
  • 00.0.0.0 → ❌ Invalid (leading zero)

Solution Development

Strategy

  1. Split the string by . to get the blocks.
  2. Check that there are exactly 4 blocks.
  3. For each block:
    • Only digits (/^\d+$/).
    • No leading zeros (except “0”).
    • Value between 0 and 255.
  4. If all checks pass, it’s a valid IPv4.

Flowchart

flowchart TD
  A["Input string"] --> B["split('.')"]
  B --> C["4 blocks?"]
  C -- No --> X["Invalid"]
  C -- Yes --> D["Validate each block"]
  D --> E["Only digits, no leading zeros"]
  E -- No --> X
  E -- Yes --> F["0 <= value <= 255?"]
  F -- No --> X
  F -- Yes --> G["Next block or valid"]
  G --> D
  D -->|All valid| Y["Valid"]

Final Code

JavaScript Implementation

function isValidIPv4(ipv4) {
  const blocks = ipv4.split('.')
  if (blocks.length !== 4)
    return false
  for (const block of blocks) {
    if (!/^\d+$/.test(block))
      return false // Only digits
    if (block.length > 1 && block[0] === '0')
      return false // Leading zero
    const num = Number(block)
    if (num < 0 || num > 255)
      return false
  }
  return true
}

Complexity Analysis

Time Complexity

The function is O(n)O(n), where nn is the length of the string:

  • split('.') scans the whole string.
  • The loop is fixed (4 blocks), each validation is O(1)O(1).

Space Complexity

It’s O(1)O(1): only 4 blocks and simple variables are used.

Edge Cases and Considerations

  • Empty or only spaces: " ", "" → ❌
  • Empty blocks: "192..1.1", "192.168..1" → ❌
  • Special characters/letters: "192.168.a.1", "192.168.1.@" → ❌
  • Negative numbers: "192.168.1.-1" → ❌
  • Leading zeros: "01.2.3.4", "192.168.01.1" → ❌
  • Incorrect number of blocks: "192.168.1", "192.168.1.1.1" → ❌
  • Out of range: "256.100.100.100", "192.168.1.300" → ❌
  • Spaces in blocks: "192.168.1. 1", " 192.168.1.1" → ❌

Reflections and Learnings

What concepts are applied?

  • String manipulation (split)
  • Regular expressions
  • Numeric conversion and validation
  • Multiple validation in a loop

Can it be optimized?

  • A global regex could be used, but the current solution is clearer and more maintainable.
  • If processing millions of IPs, you can skip number conversion if the block is already invalid.

Resources

freecodecamp daily-challenge