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Word Pattern II⚓︎

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Description⚓︎

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 non-empty 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:
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'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⚓︎

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class Solution {
private:
    bool backtracking(int pIndex, int sIndex, 
    const string& pattern, const string& s, 
    unordered_map<char, string>& patternToWord, 
    unordered_map<string, char>& wordToPattern) {

        if (pIndex == pattern.size() && sIndex == s.size()) return true;
        if (pIndex >= pattern.size() || sIndex >= s.size()) return false;

        char currentChar = pattern[pIndex];

        for (int i = sIndex; i < s.size(); i++) {
            string word = s.substr(sIndex, i - sIndex + 1);

            if ((patternToWord.count(currentChar) && patternToWord[currentChar] == word)
            || (wordToPattern.count(word) && wordToPattern[word] == currentChar)) {
                if (backtracking(pIndex + 1, i + 1, pattern, s, patternToWord, wordToPattern)) 
                    return true;
            } else if (!patternToWord.count(currentChar) && !wordToPattern.count(word)) {
                patternToWord[currentChar] = word;
                wordToPattern[word] = currentChar;

                if (backtracking(pIndex + 1, i + 1, pattern, s, patternToWord, wordToPattern))
                    return true;

                patternToWord.erase(currentChar);
                wordToPattern.erase(word);
            }
        }

        return false;
    }

public:
    bool wordPatternMatch(string pattern, string s) {
        unordered_map<char, string> patternToWord;
        unordered_map<string, char> wordToPattern;

        return backtracking(0, 0, pattern, s, patternToWord, wordToPattern);
    }
};