How to Solve Longest Word in Dictionary through Deleting

The problem requires finding the longest word in the dictionary that can be formed by deleting characters from a given string. The solution typically uses a two-pointer technique to efficiently scan through the string and check each dictionary word. If there are multiple results, return the longest word with the smallest lexicographical order.

Problem Statement

You are given a string s and a list of strings dictionary. Your task is to find the longest string in dictionary that can be formed by deleting some characters from s. If multiple words meet the criteria, return the one with the smallest lexicographical order. If no word in the dictionary can be formed, return an empty string.

The input string s and the words in dictionary consist of only lowercase English letters. The task should be solved efficiently, considering both time and space complexities, given that the string length and dictionary size can go up to 1000.

Examples

Example 1

Input: s = "abpcplea", dictionary = ["ale","apple","monkey","plea"]

Output: "apple"

Example details omitted.

Example 2

Input: s = "abpcplea", dictionary = ["a","b","c"]

Output: "a"

Example details omitted.

Constraints

• 1 <= s.length <= 1000
• 1 <= dictionary.length <= 1000
• 1 <= dictionary[i].length <= 1000
• s and dictionary[i] consist of lowercase English letters.

Solution Approach

Two-Pointer Scanning

The two-pointer approach is essential for this problem. One pointer traverses the string s while the other pointer scans through each word in the dictionary. For each word, you move the pointer on s to check if the word can be formed by deleting characters in s.

Tracking Valid Words

To efficiently track the longest valid word, maintain a variable to store the longest word found. During the two-pointer traversal, whenever a word is valid (i.e., can be formed by deleting characters), compare its length to the current longest word. If it's longer or lexicographically smaller, update the result.

Early Termination Optimization

You can optimize the solution by stopping early when a word is found that is the longest possible, reducing unnecessary checks for smaller dictionary words.

Complexity Analysis

The time complexity depends on the number of dictionary words and the length of s. For each word in the dictionary, you scan through s with two pointers, resulting in a time complexity of O(n * m), where n is the length of s and m is the average length of the dictionary words. Space complexity is O(1), since only a few pointers and variables are used.

What Interviewers Usually Probe

• Check if the candidate effectively uses two-pointer scanning to match characters between s and dictionary words.
• Look for optimization strategies, especially how they handle lexicographical order and early termination.
• Assess how the candidate handles edge cases, such as when no words can be formed from s.

Common Pitfalls or Variants

Common pitfalls

• Failing to account for lexicographical order when multiple words of the same length are valid.
• Not using the two-pointer method efficiently, leading to unnecessary scans of s.
• Overcomplicating the problem with excessive nested loops instead of leveraging simple pointer movement.

Follow-up variants

• The problem could be extended to handle uppercase letters or include non-English characters in the string and dictionary.
• A variant could involve finding the second longest valid word if the longest one has multiple occurrences.
• The task might be modified to allow for insertion or replacement of characters instead of only deletions.

How GhostInterview Helps

• GhostInterview helps by providing a step-by-step approach to solving problems using the two-pointer technique, ensuring candidates can articulate the pattern efficiently.
• The tool aids in identifying common pitfalls by guiding users to focus on optimizing the two-pointer traversal and handling edge cases correctly.
• By offering real-time feedback, GhostInterview ensures that candidates understand the intricacies of lexicographical order and early termination optimization.

Topic Pages

• Array
• Two Pointers
• String
• Sorting

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FAQ

What is the two-pointer approach in the context of the Longest Word in Dictionary through Deleting problem?

In this problem, the two-pointer approach involves using one pointer to traverse the string s and another pointer to track each dictionary word, checking if characters can be deleted to form the word.

How can I optimize the solution for the problem 'Longest Word in Dictionary through Deleting'?

You can optimize by using early termination when you find the longest valid word or by eliminating unnecessary checks for words that are shorter than the current longest valid word.

What are common mistakes when solving 'Longest Word in Dictionary through Deleting'?

Common mistakes include neglecting lexicographical order, inefficient scanning with nested loops, and not handling edge cases where no words can be formed.

How does lexicographical order affect the solution to this problem?

If multiple words can be formed from s, lexicographical order is used to return the smallest one. This ensures that among equally long valid words, the word that appears first alphabetically is returned.

Can the two-pointer approach be applied to other string-related problems?

Yes, the two-pointer approach is widely used for problems involving string matching, subsequences, or partitions, where elements need to be compared or scanned in parallel.