Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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 <= 1001 <= searchWord.length <= 10sentence consists of lowercase English letters and spaces.searchWord consists of lowercase English letters.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · String Matching
"i love eating burger" "burg"
"this problem is an easy problem" "pro"
"i am tired" "you"
counting-words-with-a-given-prefix)count-prefixes-of-a-given-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence
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;
}
}
// Accepted solution for LeetCode #1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence
func isPrefixOfWord(sentence string, searchWord string) int {
for i, s := range strings.Split(sentence, " ") {
if strings.HasPrefix(s, searchWord) {
return i + 1
}
}
return -1
}
# Accepted solution for LeetCode #1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence
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
// Accepted solution for LeetCode #1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence
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
}
}
// Accepted solution for LeetCode #1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence
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;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.