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