2024-12-02 Daily Challenge

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Posted modified:
In: DailyChallenge

Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 2

Description

Check If a Word Occurs As a Prefix of Any Word in a Sentence

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.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
vector<string> split(string &s, char spliter = ' ') {
  vector<string> result;
  for(int i = 0; i < s.length(); ++i) {
    string st;
    while(i < s.length() && s[i] != spliter) {
      st += s[i];
      i += 1;
    }
    if(st.size()) result.emplace_back(st);
  }
  return result;
}
class Solution {
public:
  int isPrefixOfWord(string sentence, string searchWord) {
    auto words = split(sentence);
    int pos = 1;
    for(const auto &word : words) {
      if(word.length() >= searchWord.length() && word.substr(0, searchWord.length()) == searchWord) return pos;
      pos += 1;
    }
    return -1;
  }
};

// Accepted
// 42/42 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 6.42 % of cpp submissions (8.3 MB)
Algorithm