How do you solve Word Squares II on LeetCode?

Word Squares II (LeetCode #3799) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Word squares depend on prefix matching.

What is the time complexity of Word Squares II?

The optimal solution for Word Squares II runs in O(n) time and uses O(n) space. The complexity is primarily due to the prefix map and backtracking, allowing efficient pruning of invalid paths.

What is the brute force solution for Word Squares II?

The brute force approach for Word Squares II is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all combinations of 4 words and check if they form a valid word square. This approach is straightforward but inefficient as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Word Squares II?

Word Squares II uses the following concepts: Array, String, Backtracking, Sorting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Word Squares II?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems beforehand.

What are common mistakes when solving Word Squares II?

Not checking all prefix conditions. Failing to handle duplicates in combinations.

#3799

Word Squares II

Medium

ArrayStringBacktrackingSortingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a backtracking approach to build word squares incrementally, ensuring at each step that the current configuration can still lead to a valid square.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix map to store words by their prefixes.
2Step 2: Use backtracking to build the word square row by row, checking prefixes against the map.
3Step 3: If a valid square is formed, add it to the result list.

solution.py20 lines

1from collections import defaultdict
2
3def wordSquares(words):
4    def backtrack(square):
5        if len(square) == 4:
6            result.append(square[:])
7            return
8        prefix = ''.join([square[i][len(square)] for i in range(len(square))])
9        for word in prefix_map[prefix]:
10            square.append(word)
11            backtrack(square)
12            square.pop()
13
14    prefix_map = defaultdict(list)
15    for word in words:
16        for i in range(1, 5):
17            prefix_map[word[:i]].append(word)
18    result = []
19    backtrack([])
20    return sorted(result)

ℹ

Complexity note: The complexity is primarily due to the prefix map and backtracking, allowing efficient pruning of invalid paths.

1Word squares depend on prefix matching.
2Backtracking allows efficient exploration of valid configurations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.