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Problem: Shortest Word Distance II

med
30 min
Explore how to create a WordDistance class that initializes with an array of strings and efficiently computes the shortest distance between any two different words. Understand the problem constraints and implement methods to handle multiple queries, improving your ability to design specialized data structures.

Statement

Design a data structure that takes in an array of strings and efficiently computes the shortest distance between any two different strings in the array.

Implement the WordDistance class:

  • WordDistance(String[] words_dict): Initializes the object with an array of strings.

  • int shortest(String word1, String word2): Returns the shortest distance between word1 and word2 in the array of strings.

Constraints:

  • 11 \leq words_dict.length 103\leq10^3

  • 11 \leq words_dict[i].length 10\leq 10

  • words_dict[i] consists of lowercase English letters

  • word1 and word2 are in words_dict

  • word1 != word2

  • At most, 10001000 calls will be made to the shortest

Tap here to switch tabs
Problem
Ask
Submissions

Problem: Shortest Word Distance II

med
30 min
Explore how to create a WordDistance class that initializes with an array of strings and efficiently computes the shortest distance between any two different words. Understand the problem constraints and implement methods to handle multiple queries, improving your ability to design specialized data structures.

Statement

Design a data structure that takes in an array of strings and efficiently computes the shortest distance between any two different strings in the array.

Implement the WordDistance class:

  • WordDistance(String[] words_dict): Initializes the object with an array of strings.

  • int shortest(String word1, String word2): Returns the shortest distance between word1 and word2 in the array of strings.

Constraints:

  • 11 \leq words_dict.length 103\leq10^3

  • 11 \leq words_dict[i].length 10\leq 10

  • words_dict[i] consists of lowercase English letters

  • word1 and word2 are in words_dict

  • word1 != word2

  • At most, 10001000 calls will be made to the shortest