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

January LeetCoding Challenge 1

Description

Word Pattern

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"
Output: true

Example 2:

Input: pattern = "abba", s = "dog cat cat fish"
Output: false

Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false

Constraints:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letters.
• 1 <= s.length <= 3000
• s contains only lowercase English letters and spaces ' '.
• s does not contain any leading or trailing spaces.
• All the words in s are separated by a single space.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  int countWords(string &s) {
    bool isSpace = true;
    int result = 0;
    for(auto c : s) {
      result += (c != ' ') * isSpace;
      isSpace = c == ' ';
    }
    return result;
  }
  string nextWord(string &s, int &index) {
    while(s[index] == ' ') index += 1;
    int len = 0;
    while(index + len < s.length() && s[index + len] != ' ') len += 1;
    index += len;
    return s.substr(index - len, len);
  }
public:
  bool wordPattern(string pattern, string s) {
    if(pattern.length() != countWords(s)) return false;
    unordered_map<char, string> mp;
    unordered_map<string, int> cnt;
    int pos = 0;
    for(auto c : pattern) {
      auto word = nextWord(s, pos);
      if(mp.count(c) && mp[c] != word) return false;
      if(!mp.count(c)){
        mp[c] = word;
        cnt[word] += 1;
      } 
    }
    for(auto [_word, c] : cnt) if(c > 1) return false;
    return true;
  }
};

// Accepted
// 36/36 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 82.42 % of cpp submissions (6.3 MB)