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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.
1import java.util.HashSet;
2import java.util.Arrays;
3
4public class Solution {
5 public static
In Java, the solution employs a HashSet
to keep track of valid prefixes. Sorting is done via a custom comparator that sorts by length and lexicographical order. This helps in efficiently deciding which word is substantiated as a valid word as we iterate through the sorted array.
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.