Solution Explanation

The problem asks to find the longest word in a given dictionary that can be built one character at a time by other words in the same dictionary. If multiple such words exist, the lexicographically smallest one should be returned.

The solutions presented utilize a combination of techniques to efficiently solve this problem:

1. Trie (Not explicitly used but implicitly leveraged in some solutions): While not explicitly implemented as a Trie data structure, the solutions implicitly leverage the Trie concept. A Trie is a tree-like data structure ideal for storing and searching prefixes. The solutions efficiently check if prefixes of a word exist in the dictionary by iterating through the prefixes and checking their presence in a set. This is fundamentally the same operation as traversing a Trie.

2. Hash Table (Set): A hash set (or simply a set in Python) is used to store the words in the dictionary. This allows for efficient O(1) average-time lookup of whether a word or prefix exists in the dictionary.

3. Sorting (In some solutions): Some solutions utilize sorting to optimize the search. By sorting the words by length (descending) and then lexicographically (ascending), the algorithm can potentially find a solution faster. It checks longer words first, and if a longest word is found that satisfies the condition, it immediately returns the result, without checking other potentially longer words.

4. Iteration and Prefix Checking: The core logic of the solution involves iterating through each word in the dictionary. For each word, it iterates through its prefixes. The solution checks if each prefix is present in the dictionary. If all prefixes are present, then the word satisfies the condition and is considered a candidate.

Time Complexity Analysis:

• Without sorting: The time complexity is dominated by the nested loops. The outer loop iterates through each word (O(n), where n is the number of words). The inner loop iterates through the prefixes of each word (O(m), where m is the maximum length of a word). Therefore, the overall time complexity is O(n*m). However, the set lookups are O(1) on average, making this dominant factor.

• With sorting: The sorting step adds O(n log n) time complexity (where n is number of words). The nested loop still remains, but by pre-sorting, the algorithm can be more efficient because the solution can potentially exit early, avoiding unnecessary checks. In the worst case, the complexity remains O(n*m), but in the best case, it might be much faster than the unsorted version if the longest word satisfying the conditions is one of the first elements after sorting.

Space Complexity Analysis:

The space complexity is dominated by the space used to store the hash set (or set), which is O(n) where n is the number of words in the dictionary. The space used by other variables is constant, making O(n) the overall space complexity.

Code Explanation (Python3 Example):

The Python solution demonstrates the approach clearly:

class Solution:
    def longestWord(self, words: List[str]) -> str:
        cnt, ans = 0, ''  # Initialize variables to track length and the result
        s = set(words)     # Create a set for efficient word lookups
 
        for w in s:        # Iterate through each word in the set
            n = len(w)     # Get the length of the word
            if all(w[:i] in s for i in range(1, n)):  # Check if all prefixes exist in the set
                if cnt < n:  # If this word is longer than the current longest
                    cnt, ans = n, w # Update the length and the longest word
                elif cnt == n and w < ans:  # If same length, compare lexicographically
                    ans = w             # Update the longest word if lexicographically smaller
        return ans         # Return the longest word found

The other language implementations follow a similar logic, differing primarily in syntax and data structure usage. The core algorithm remains consistent across all the implementations.