# 291. Word Pattern II

Given a `pattern` and a string `s`, return `true` *if* `s` ***matches** the* `pattern`*.*

A string `s` **matches** a `pattern` if there is some **bijective mapping** of single characters to strings such that if each character in `pattern` is replaced by the string it maps to, then the resulting string is `s`. A **bijective mapping** means that no two characters map to the same string, and no character maps to two different strings.

 

**Example 1:**

```
Input: pattern = "abab", s = "redblueredblue"
Output: true
Explanation: One possible mapping is as follows:
'a' -> "red"
'b' -> "blue"
```

**Example 2:**

```
Input: pattern = "aaaa", s = "asdasdasdasd"
Output: true
Explanation: One possible mapping is as follows:
'a' -> "asd"
```

**Example 3:**

```
Input: pattern = "aabb", s = "xyzabcxzyabc"
Output: false
```

 

**Constraints:**

* `1 <= pattern.length, s.length <= 20`
* `pattern` and `s` consist of only lowercase English letters.

### **Solution:**

[**https://www.jiuzhang.com/problem/word-pattern-ii/**](https://www.jiuzhang.com/problem/word-pattern-ii/)\ <br>

```
public class Solution {
    /**
     * @param pattern: a string,denote pattern string
     * @param str: a string, denote matching string
     * @return: a boolean
     */
    public boolean wordPatternMatch(String pattern, String str) {
        Map<Character, String> map = new HashMap<>();
        Set<String> set = new HashSet<>();
        return match(pattern, str, map, set);
    }
    
    private boolean match(String pattern,
                          String str,
                          Map<Character, String> map,
                          Set<String> set) {
        if (pattern.length() == 0) {
            return str.length() == 0;
        }
        
        Character c = pattern.charAt(0);
        if (map.containsKey(c)) {
            if (!str.startsWith(map.get(c))) {
                return false;
            }
            
            return match(
                pattern.substring(1),
                str.substring(map.get(c).length()),
                map,
                set
            );
        }
        
        for (int i = 0; i < str.length(); i++) {
            String word = str.substring(0, i + 1);
            if (set.contains(word)) {
                continue;
            }
            map.put(c, word);
            set.add(word);
            if (match(pattern.substring(1),
                      str.substring(i + 1),
                      map,
                      set)) {
                return true;              
            }
            set.remove(word);
            map.remove(c);
        }
        
        return false;
    }
}
```


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