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# Word Pattern LeetCode Solution

## Problem – Word Pattern LeetCode Solution

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.

## Word Pattern LeetCode Solution in Java

``````public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (words.length != pattern.length())
return false;
Map index = new HashMap();
for (Integer i=0; i<words.length; ++i)
if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
return false;
return true;
}
``````

## Word Pattern LeetCode Solution in C++

``````bool wordPattern(string pattern, string str) {
map<char, int> p2i;
map<string, int> w2i;
istringstream in(str);
int i = 0, n = pattern.size();
for (string word; in >> word; ++i) {
if (i == n || p2i[pattern[i]] != w2i[word])
return false;
p2i[pattern[i]] = w2i[word] = i + 1;
}
return i == n;
}
``````

## Word Pattern LeetCode Solution in Python

``````def wordPattern(self, pattern, str):
f = lambda s: map({}.setdefault, s, range(len(s)))
return f(pattern) == f(str.split())
``````
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