LeetCode 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence

Description

https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/

Given a sentence that consists of some words separated by a single space, and a searchWord.

You have to check if searchWord is a prefix of any word in sentence.

Return the index of the word in sentence where searchWord is a prefix of this word (1-indexed).

If searchWord is a prefix of more than one word, return the index of the first word (minimum index). If there is no such word return -1.

prefix of a string S is any leading contiguous substring of S.

Example 1:

Input: sentence = "i love eating burger", searchWord = "burg"
Output: 4
Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.

Example 2:

Input: sentence = "this problem is an easy problem", searchWord = "pro"
Output: 2
Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.

Example 3:

Input: sentence = "i am tired", searchWord = "you"
Output: -1
Explanation: "you" is not a prefix of any word in the sentence.

Example 4:

Input: sentence = "i use triple pillow", searchWord = "pill"
Output: 4

Example 5:

Input: sentence = "hello from the other side", searchWord = "they"
Output: -1

Constraints:

  • 1 <= sentence.length <= 100
  • 1 <= searchWord.length <= 10
  • sentence consists of lowercase English letters and spaces.
  • searchWord consists of lowercase English letters.

Explanation

Break the sentence to word list and check the searchWord is the prefix for which words.

Python Solution

class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        words = sentence.split()
        
        words_dict = {}
        
        for i, word in enumerate(words):
            words_dict[i] = word
            
        for key, value in words_dict.items():
            if value.startswith(searchWord):
                return key + 1
            
            
        return -1            
        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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