Input
["WordFilter", "f"]
[[["apple"]], ["a", "e"]]
Output
[null, 0]
Explanation
WordFilter wordFilter = new WordFilter(["apple"]);
wordFilter.f("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = "e".

Solution Explanation for Prefix and Suffix Search

This problem requires designing a data structure that efficiently searches for words with a given prefix and suffix. Two main solutions are presented: a hash table-based approach and a Trie-based approach.

Solution 1: Hash Table Approach

This approach uses a hash table (dictionary in Python) to store all possible prefix-suffix combinations and their corresponding word indices.

Algorithm:

1. Initialization (__init__ or constructor):

   • Create a hash table d.
   • Iterate through the input words list.
   • For each word w:
     • Iterate through all possible prefixes (from empty string to the entire word).
     • For each prefix a, iterate through all possible suffixes (from the entire word to the empty string).
     • Store the pair (a, b) as the key in d, and the index k of the word as the value.

2. Search (f):

   • Given a pref and suff, construct the key (pref, suff).
   • Return the value associated with this key in d (the index). If the key is not found, return -1.

Time Complexity:

• Initialization: O(N * L^2), where N is the number of words and L is the maximum length of a word. This is because we iterate through all prefixes and suffixes for each word.
• Search (f): O(1) on average, as hash table lookups are typically O(1). In the worst case, it could be O(N) if there are many hash collisions (though this is unlikely with a good hash function).

Space Complexity: O(N * L^2) in the worst case, to store all possible prefix-suffix combinations.

Solution 2: Trie Approach

This approach uses two Tries: one for prefixes and one for suffixes.

Algorithm:

1. Initialization (__init__ or constructor):

   • Create two Tries, p (for prefixes) and s (for suffixes).
   • Iterate through the input words list.
   • For each word w and its index i:
     • Insert w into p.
     • Insert the reversed w into s (to efficiently search suffixes).

2. Search (f):

   • Given pref and suff, search for pref in p and the reversed suff in s. This will give lists of indices (a and b).
   • If either list is empty, return -1.
   • Iterate through a and b from the end, finding the largest index that is present in both lists.

Time Complexity:

• Initialization: O(N * L), where N is the number of words and L is the maximum length of a word. Insertion into a Trie takes O(L) time.
• Search (f): O(L) on average. Searching in a Trie takes O(L) time. The two-pointer comparison afterwards is at most O(N), but in practice, it's often much smaller since we only compare until the common element is found.

Space Complexity: O(N * L) in the worst case, as the Tries store all characters of all words.

Which Solution is Better?

Solution 2 (Trie-based) is generally preferred because it has a lower time complexity for both initialization and search, especially for larger datasets. While Solution 1 has a simpler implementation, its quadratic time complexity for initialization can become a bottleneck. The space complexity is also improved in Solution 2. The Trie efficiently handles the overlapping prefixes and suffixes.