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.
1 2 3 4 5 6 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|