Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #408: Valid Word Abbreviation

In this guide, we solve Leetcode #408 Valid Word Abbreviation in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

A string can be abbreviated by replacing any number of non-adjacent, non-empty substrings with their lengths. The lengths should not have leading zeros.

Quick Facts

  • Difficulty: Easy
  • Premium: Yes
  • Tags: Two Pointers, String

Intuition

The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.

Moving one pointer at a time keeps the invariant intact and avoids nested loops.

Approach

Place pointers at the left and right ends and move them based on the comparison or target condition.

This yields a clean linear pass after any required sorting.

Steps:

  • Set left and right pointers.
  • Move a pointer based on the condition.
  • Update the best answer while scanning.

Example

Input: word = "internationalization", abbr = "i12iz4n" Output: true Explanation: The word "internationalization" can be abbreviated as "i12iz4n" ("i nternational iz atio n").

Python Solution

class Solution: def validWordAbbreviation(self, word: str, abbr: str) -> bool: m, n = len(word), len(abbr) i = j = x = 0 while i < m and j < n: if abbr[j].isdigit(): if abbr[j] == "0" and x == 0: return False x = x * 10 + int(abbr[j]) else: i += x x = 0 if i >= m or word[i] != abbr[j]: return False i += 1 j += 1 return i + x == m and j == n

Complexity

The time complexity is O(m+n)O(m + n)O(m+n), where mmm and nnn are the lengths of the string wordwordword and the string abbrabbrabbr respectively. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy