Leetcode #425: Word Squares
In this guide, we solve Leetcode #425 Word Squares 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 unique strings words, return all the word squares you can build from words. The same word from words can be used multiple times.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Trie, Array, String, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: words = ["area","lead","wall","lady","ball"]
Output: [["ball","area","lead","lady"],["wall","area","lead","lady"]]
Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 26
self.v = []
def insert(self, w, i):
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.v.append(i)
def search(self, w):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return []
node = node.children[idx]
return node.v
class Solution:
def wordSquares(self, words: List[str]) -> List[List[str]]:
def dfs(t):
if len(t) == len(words[0]):
ans.append(t[:])
return
idx = len(t)
pref = [v[idx] for v in t]
indexes = trie.search(''.join(pref))
for i in indexes:
t.append(words[i])
dfs(t)
t.pop()
trie = Trie()
ans = []
for i, w in enumerate(words):
trie.insert(w, i)
for w in words:
dfs([w])
return ans
Complexity
The time complexity is Exponential (worst case). The space complexity is O(depth).
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.