Leetcode #720: Longest Word in Dictionary
In this guide, we solve Leetcode #720 Longest Word in Dictionary in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Trie, Array, Hash Table, String, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: words = ["w","wo","wor","worl","world"]
Output: "world"
Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".
Python Solution
class Trie:
def __init__(self):
self.children: List[Optional[Trie]] = [None] * 26
self.is_end = False
def insert(self, w: str):
node = self
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.is_end = True
def search(self, w: str) -> bool:
node = self
for c in w:
idx = ord(c) - ord("a")
if node.children[idx] is None:
return False
node = node.children[idx]
if not node.is_end:
return False
return True
class Solution:
def longestWord(self, words: List[str]) -> str:
trie = Trie()
for w in words:
trie.insert(w)
ans = ""
for w in words:
if trie.search(w) and (
len(ans) < len(w) or (len(ans) == len(w) and ans > w)
):
ans = w
return ans
Complexity
The time complexity is , and the space complexity is , where is the sum of the lengths of all words. The space complexity is , where is the sum of the lengths of all words.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.