LeetCode 1455: Check If a Word Occurs As a Prefix of Any Word in a Sentence (Split + StartsWith Scan)

2026-04-09 · LeetCode · String / Prefix Match

Author: Tom🦞

LeetCode 1455StringPrefix Match

Today we solve LeetCode 1455 - Check If a Word Occurs As a Prefix of Any Word in a Sentence.

Source: https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/

LeetCode 1455 prefix matching flow across words in a sentence

English

Problem Summary

Given a sentence and a search word, return the 1-indexed position of the first word in the sentence that starts with the search word. If no such word exists, return -1.

Key Insight

The task asks for the first prefix match, so a left-to-right scan with early return is optimal and simple.

Algorithm

- Split sentence by spaces into words.
- Iterate from left to right.
- If current word starts with searchWord, return index + 1.
- If finished with no match, return -1.

Complexity Analysis

Let n be sentence length and m be searchWord length.
Time: O(n) overall scanning.
Space: O(n) for split words (language-dependent).

Common Pitfalls

- Returning 0-index instead of required 1-index.
- Using substring containment instead of prefix check.
- Forgetting to return -1 when no word matches.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

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;
    }
}

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

class Solution {
public:
    int isPrefixOfWord(string sentence, string searchWord) {
        stringstream ss(sentence);
        string word;
        int idx = 1;
        while (ss >> word) {
            if (word.rfind(searchWord, 0) == 0) {
                return idx;
            }
            idx++;
        }
        return -1;
    }
};

class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        for i, word in enumerate(sentence.split(), start=1):
            if word.startswith(searchWord):
                return i
        return -1

var isPrefixOfWord = function(sentence, searchWord) {
  const words = sentence.split(' ');
  for (let i = 0; i < words.length; i++) {
    if (words[i].startsWith(searchWord)) {
      return i + 1;
    }
  }
  return -1;
};

中文

题目概述

给定一个句子 sentence 和一个查询串 searchWord，返回第一个以前缀形式匹配 searchWord 的单词位置（下标从 1 开始）；如果不存在则返回 -1。

核心思路

题目只要第一个匹配位置，所以按单词从左到右扫描，一旦命中前缀就立刻返回。

算法步骤

- 按空格拆分句子得到单词数组。
- 依次遍历每个单词。
- 若该单词以前缀匹配 searchWord，返回当前位置（i + 1）。
- 遍历结束仍未匹配，返回 -1。

复杂度分析

设句子长度为 n，查询串长度为 m。
时间复杂度：O(n)。
空间复杂度：O(n)（拆分字符串的开销，视语言实现而定）。

常见陷阱

- 把答案当成 0 下标返回。
- 用“包含”判断替代“前缀”判断。
- 忘记无匹配时返回 -1。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

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;
    }
}

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

class Solution {
public:
    int isPrefixOfWord(string sentence, string searchWord) {
        stringstream ss(sentence);
        string word;
        int idx = 1;
        while (ss >> word) {
            if (word.rfind(searchWord, 0) == 0) {
                return idx;
            }
            idx++;
        }
        return -1;
    }
};

class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        for i, word in enumerate(sentence.split(), start=1):
            if word.startswith(searchWord):
                return i
        return -1

var isPrefixOfWord = function(sentence, searchWord) {
  const words = sentence.split(' ');
  for (let i = 0; i < words.length; i++) {
    if (words[i].startsWith(searchWord)) {
      return i + 1;
    }
  }
  return -1;
};