Word Search II

Problem

Given an m x n board of characters and a list of strings words, return all words on the board.

Examples

The brute-force approach

For each word, run Word Search I (DFS with backtracking). Return all words that return true.

result = []
for word in words:
    if wordSearch(board, word):
        result.append(word)
return result

O(words × m × n × 4^L) where L = max word length. You repeat the full board search for every word independently. Common prefixes are re-explored from scratch each time.

Spotting the pattern

This is a Trie problem. The key question to ask yourself:

How does searching all words simultaneously (via Trie) reduce redundant work compared to searching one word at a time?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.