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

Description

Given a sentence that consists of some words separated by a single space, and a searchWord, check if searchWord is a prefix of any word in sentence.

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

 

Constraints:

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

Solutions

Solution 1: String Splitting

We split \(\textit{sentence}\) by spaces into \(\textit{words}\), then iterate through \(\textit{words}\) to check if \(\textit{words}[i]\) is a prefix of \(\textit{searchWord}\). If it is, we return \(i+1\). If the iteration completes and no words satisfy the condition, we return \(-1\).

The time complexity is \(O(m \times n)\), and the space complexity is \(O(m)\). Here, \(m\) and \(n\) are the lengths of \(\textit{sentence}\) and \(\textit{searchWord}\), respectively.

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class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        for i, s in enumerate(sentence.split(), 1):
            if s.startswith(searchWord):
                return i
        return -1

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

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class Solution {
public:
    int isPrefixOfWord(string sentence, string searchWord) {
        stringstream ss(sentence);
        string s;
        for (int i = 1; ss >> s; ++i) {
            if (s.find(searchWord) == 0) {
                return i;
            }
        }
        return -1;
    }
};

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func isPrefixOfWord(sentence string, searchWord string) int {
    for i, s := range strings.Split(sentence, " ") {
        if strings.HasPrefix(s, searchWord) {
            return i + 1
        }
    }
    return -1
}

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function isPrefixOfWord(sentence: string, searchWord: string): number {
    const ss = sentence.split(/\s/);
    const n = ss.length;
    for (let i = 0; i < n; i++) {
        if (ss[i].startsWith(searchWord)) {
            return i + 1;
        }
    }
    return -1;
}

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impl Solution {
    pub fn is_prefix_of_word(sentence: String, search_word: String) -> i32 {
        let ss = sentence.split_whitespace().collect::<Vec<&str>>();
        for i in 0..ss.len() {
            if ss[i].starts_with(&search_word) {
                return (i + 1) as i32;
            }
        }
        -1
    }
}

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class Solution {
    /**
     * @param String $sentence
     * @param String $searchWord
     * @return Integer
     */
    function isPrefixOfWord($sentence, $searchWord) {
        $words = explode(' ', $sentence);
        for ($i = 0; $i < count($words); ++$i) {
            if (strpos($words[$i], $searchWord) === 0) {
                return $i + 1;
            }
        }
        return -1;
    }
}

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