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

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Description

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.

A 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.

Solutions

Python3

class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        words = sentence.split(' ')
        i, n = 0, len(words)
        while i < n:
            word = words[i]
            if word[: len(searchWord)] == searchWord:
                return i + 1
            i += 1
        return -1

Java

class Solution {
    public int isPrefixOfWord(String sentence, String searchWord) {
        String[] words = sentence.split(" ");
        int i = 0, n = words.length;
        for (; i < n; ++i) {
            if (words[i].indexOf(searchWord) == 0) {
                return i + 1;
            }
        }
        return -1;
    }
}

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