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This approach involves sorting the words array. By sorting, words that could potentially be built upon would appear earlier in the list. After sorting, we use a set to keep track of words that can be built character by character.
The process is as follows:
This is efficient due to the pre-sorting which allows easy checking and ensures we are building words correctly by the constraints given.
Time complexity: O(N log N + N * M), where N is the number of words and M is the average length of the words. Sorting takes O(N log N) and checking prefixes takes O(N * M).
Space complexity: O(N * M), where M is the length of the longest word.
1function longestWord(words) {
2 words.sort((a, b) => a.length === b.length ? (a < b ? -1 :
In JavaScript, the solution leverages Set for quick prefix checks. Sorting is handled with Array sort method using a comparator that first checks length and then lexicographical order. This allows direct evaluation of the validity of each word in sequence.
This approach uses a Trie (prefix tree) to store the words. By utilizing this data structure, we aim to verify if a word can be constructed by progressively building on its prefixes.
The steps are:
This combines efficient storage and lookup, with Prefix trees naturally lending themselves to this problem due to their structure.
Time complexity: O(N * M), where creating the Trie and checking word validity both operate over each character of every word.
Space complexity: O(N * M) for the Trie data structure.
1class TrieNode:
2
This Python solution involves first inserting all words into a Trie structure and then checking each sorted word to verify that every prefix exists within the Trie. The check is aided by the use of an end_of_word flag to determine the presence of complete valid prefixes every step of the way.