Leetcode #472: Concatenated Words
In this guide, we solve Leetcode #472 Concatenated Words 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 (without duplicates), return all the concatenated words in the given list of words. A concatenated word is defined as a string that is comprised entirely of at least two shorter words (not necessarily distinct) in the given array.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Trie, Array, String, Dynamic Programming, Sorting
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 26
self.is_end = False
def insert(self, w):
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
class Solution:
def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
def dfs(w):
if not w:
return True
node = trie
for i, c in enumerate(w):
idx = ord(c) - ord('a')
if node.children[idx] is None:
return False
node = node.children[idx]
if node.is_end and dfs(w[i + 1 :]):
return True
return False
trie = Trie()
ans = []
words.sort(key=lambda x: len(x))
for w in words:
if dfs(w):
ans.append(w)
else:
trie.insert(w)
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.