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.
class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        sentence = sentence.split(' ')
        a = []
        for i in range(len(sentence)):
            if len(sentence[i]) >= len(searchWord):
                if sentence[i][0:len(searchWord)] == searchWord:
                    a.append(i)
        if a:
            return min(a)+1
        else:
            return -1
class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        sentence = sentence.split(' ')
        a = []
        for i in range(len(sentence)):
            if len(sentence[i]) >= len(searchWord):
                if sentence[i][0:len(searchWord)] == searchWord:
                    a.append(i)
        if a:
            return min(a)+1
        else:
            return -1