# 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).