110 lines
2.1 KiB
Markdown
110 lines
2.1 KiB
Markdown
# Leetcode Word-Pattern
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2022-09-05 19:27
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> ##### Data structures:
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>
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> #DS #string
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty_easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Links:
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- [Link to problem]()
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---
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### Problem
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Given a `pattern` and a string `s`, find if `s` follows the same pattern.
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Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
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#### Examples
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**Example 1:**
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**Input:** pattern = "abba", s = "dog cat cat dog"
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**Output:** true
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**Example 2:**
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**Input:** pattern = "abba", s = "dog cat cat fish"
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**Output:** false
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**Example 3:**
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**Input:** pattern = "aaaa", s = "dog cat cat dog"
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**Output:** false
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#### Constraints
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- `1 <= pattern.length <= 300`
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- `pattern` contains only lower-case English letters.
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- `1 <= s.length <= 3000`
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- `s` contains only lowercase English letters and spaces `' '`.
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- `s` **does not contain** any leading or trailing spaces.
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- All the words in `s` are separated by a **single space**.
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### Thoughts
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> [!summary]
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> This is a #string operation problem.
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The main part is using two hash tables, to check for
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mismatches, and using `istringstream` to read from string
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Two hash tables:
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- One is used to check for s has one and only bound word
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- The other is used to check that the word is only stored
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once
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Using one hash table is one way, which makes one way
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unchecked.
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### Solution
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```cpp
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class Solution {
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public:
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bool wordPattern(string pattern, string s) {
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unordered_map<string, int> stoi;
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unordered_map<char, int> ctoi;
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// convert string to stream
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istringstream input(s);
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string word;
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int n = pattern.size();
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int id;
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for (id = 0; input >> word; id++) {
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// cout<<word<<'\n';
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if (id == n || stoi[word] != ctoi[pattern[id]]) {
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// pattern OOB, or mismatch.
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return false;
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}
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// to prevent id == 0 is the default value of map
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stoi[word] = ctoi[pattern[id]] = id + 1;
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}
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if (id == n) {
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return true;
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} else {
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// check for lingering word in s
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return false;
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}
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}
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};
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```
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