LEETCODE 408 C SOLUTION:

Welcome to this detailed walkthrough of LeetCode Problem 408, "Valid Word Abbreviation." In this video, we will delve into the problem statement, explore various strategies to solve it, and provide an in-depth explanation of the most efficient approach. This video is designed for programmers at all levels, whether you're a beginner looking to understand the basics or an experienced coder seeking to refine your skills.
Introduction
LeetCode Problem 408, "Valid Word Abbreviation," challenges us to determine whether a given abbreviation is a valid abbreviation of a given word. This problem tests our understanding of string manipulation and the ability to handle different edge cases effectively.
Problem Statement
The problem provides us with a word and an abbreviation, and our task is to check if the abbreviation correctly represents the word. An abbreviation is valid if it follows the rules:
Each number in the abbreviation represents the number of characters to skip in the word.
Any alphabetic character in the abbreviation must match the corresponding character in the word.
For example, given the word "internationalization" and the abbreviation "i12iz4n," the abbreviation is valid because:
'i' matches 'i'.
'12' skips 12 characters.
'i' matches 'i'.
'z' matches 'z'.
'4' skips 4 characters.
'n' matches 'n'.
Approach to Solve the Problem
To solve this problem, we need a method that traverses both the word and the abbreviation simultaneously, verifying the validity of the abbreviation at each step. Here is a step-by-step breakdown of our approach:
Initialize Pointers: Start with two pointers, one for the word and one for the abbreviation.
Traverse Both Strings: Iterate through both strings, comparing characters or handling numbers that represent skips in the word.
Handle Edge Cases: Ensure proper handling of edge cases, such as leading zeros in the abbreviation or mismatched characters.
Detailed Explanation
Step 1: Initialize Pointers
We initialize two pointers, i for the word and j for the abbreviation. These pointers will help us track our position in both strings.
Step 2: Traverse Both Strings
As we traverse the abbreviation, we perform the following checks:
If the current character in the abbreviation is a letter, it must match the corresponding character in the word.
If the current character in the abbreviation is a digit, we calculate the number it represents and move the pointer in the word accordingly.
Step 3: Handle Edge Cases
Consider the following edge cases:
Leading Zeros: The abbreviation should not contain leading zeros, as these are invalid.
End of Strings: Ensure that both pointers reach the end of their respective strings simultaneously for the abbreviation to be valid.
Example Walkthrough
Let's walk through an example to illustrate our approach:
Example 1:
Word: "internationalization"
Abbreviation: "i12iz4n"
Steps:
i matches i.
12 means we skip the next 12 characters: "nternationaliza".
i matches i.
z matches z.
4 means we skip the next 4 characters: "atio".
n matches n.
Both pointers reach the end of their respective strings, so the abbreviation is valid.
Example 2:
Word: "apple"
Abbreviation: "a2e"
Steps:
a matches a.
2 means we skip the next 2 characters: "pp".
e matches e.
Both pointers reach the end of their respective strings, so the abbreviation is valid.
Common Pitfalls
Ignoring Leading Zeros: Ensure that the abbreviation does not contain numbers with leading zeros, as these are invalid.
Mismatch Handling: Properly handle cases where the characters do not match or the skips do not align with the word length.
End of Strings: Ensure both pointers reach the end of their respective strings to confirm the abbreviation's validity.
Optimization
While our approach is efficient, we can further optimize by avoiding unnecessary comparisons and directly handling valid and invalid cases. This ensures our solution runs in linear time relative to the length of the word and the abbreviation.
Conclusion
LeetCode Problem 408, "Valid Word Abbreviation," is an excellent problem for practicing string manipulation and understanding the nuances of string traversal. By carefully handling edge cases and ensuring our pointers are managed correctly, we can efficiently determine the validity of an abbreviation.
We hope this video has provided you with a clear and thorough understanding of the problem and its solution. If you found this video helpful, please like, subscribe, and share it with others who might benefit from it. Don't forget to hit the bell icon to receive notifications for our latest videos.
Further Practice
For further practice, try implementing the solution yourself and testing it with different examples. Consider edge cases and try to optimize your solution for better performance