notes/OJ notes/pages/Leetcode Word-Pattern.md

Leetcode Word-Pattern

2022-09-05 19:27

Data structures:

#DS #string

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Links:

• Link to problem

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Problem

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.

Examples

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.

Thoughts

[!summary] This is a #string operation problem.

The main part is using two hash tables, to check for mismatches, and using istringstream to read from string

Two hash tables:

• One is used to check for s has one and only bound word
• The other is used to check that the word is only stored once

Using one hash table is one way, which makes one way unchecked.

Solution

class Solution {
public:
  bool wordPattern(string pattern, string s) {
    unordered_map<string, int> stoi;
    unordered_map<char, int> ctoi;
    // convert string to stream
    istringstream input(s);
    string word;
    int n = pattern.size();
    int id;

    for (id = 0; input >> word; id++) {
      // cout<<word<<'\n';
      if (id == n || stoi[word] != ctoi[pattern[id]]) {
        // pattern OOB, or mismatch.
        return false;
      }

      // to prevent id == 0 is the default value of map
      stoi[word] = ctoi[pattern[id]] = id + 1;
    }

    if (id == n) {
      return true;
    } else {
      // check for lingering word in s
      return false;
    }
  }
};