Given a dictionary of words, find all strings that matches the given pattern where every character in the pattern is uniquely mapped to a character in the dictionary.
Input: dict = ["abb", "abc", "xyz", "xyy"]; pattern = "foo" Output: [xyy abb] Explanation: xyy and abb have same character at index 1 and 2 like the pattern Input: dict = ["abb", "abc", "xyz", "xyy"]; pat = "mno" Output: [abc xyz] Explanation: abc and xyz have all distinct characters, similar to the pattern Input: dict = ["abb", "abc", "xyz", "xyy"]; pattern = "aba" Output:  Explanation: Pattern has same character at index 0 and 2. No word in dictionary follows the pattern. Input: dict = ["abab", "aba", "xyz", "xyx"]; pattern = "aba" Output: [aba xyx] Explanation: aba and xyx have same character at index 0 and 2 like the pattern
The idea is to encode the pattern in such a way that any word from the dictionary that matches the pattern will have same hash as that of the pattern after encoding. We iterate through all words in dictionary one by one and print the words that have same hash as that of the pattern.
Below is the implementation of above idea –
The idea is to traverse both the pattern and word in parallel and for every first occurrence of any character in pattern we map it to the corresponding character present in word, and for any subsequent occurrence of the same character in pattern, there must me the same character in word which was earlier mapped.
For first occurrence of any character in pattern, we do ch[pattern[i]] = word[i].
and subsequent occurrence of any character must satisfy ch[pattern[i]] == word[i].
Below is the implementation of the above approach:
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