Problem

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

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Examples

Example 1:

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Input:
board = [
	["o","a","a","n"],
	["e","t","a","e"],
	["i","h","k","r"],
	["i","f","l","v"]
], 
words = ["oath","pea","eat","rain"]

Output: 
["eat","oath"]

Similar Problems

This is a follow up: Word Search 1 - Find if word exists

Boggle Solver

Boggle is a game played on a 4 x 4 grid of letters. The goal is to find as many words as possible that can be formed by a sequence of adjacent letters in the grid, using each cell at most once. Given a game board and a dictionary of valid words, implement a Boggle solver.

Key differences:

  • Movement: Boggle allows 8 directions (including diagonals), Word Search II uses 4 directions
  • Grid size: Boggle is traditionally 4x4, Word Search II can be any m×n
  • Word source: Boggle uses a dictionary, Word Search II uses a specific word list
  • Minimum length: Boggle typically requires 3+ character words
Examples
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G I Z E
U E K S  →  Can find: GEEK, QUIZ, SEEK, GUY, ZUKES
Q S E R
A B C D

Solution

Method 1 - Backtracking and DFS

Similar to Word Search 1 - Find if word exists,this problem can be solved by DFS. However, this solution exceeds time limit.

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public List<String> findWords(char[][] board, String[] words) {
	List<String> ans = new ArrayList<String> ();

	int m = board.length;
	int n = board[0].length;

	for (String word: words) {
		boolean flag = false;
		for (int i = 0; i<m; i++) {
			for (int j = 0; j<n; j++) {
				char[][] newBoard = new char[m][n];
				for (int x = 0; x<m; x++) {
					for (int y = 0; y<n; y++) {
						newBoard[x][y] = board[x][y];
					}
				}

				if (dfs(newBoard, word, i, j, 0)) {
					flag = true;
				}
			}
		}
		if (flag) {
			ans.add(word);
		}
	}

	return ans;
}

public boolean dfs(char[][] board, String word, int i, int j, int k) {
	int m = board.length;
	int n = board[0].length;

	if (i<0 || j<0 || i >= m || j >= n || k > word.length() - 1) {
		return false;
	}

	if (board[i][j] == word.charAt(k)) {
		char temp = board[i][j];
		board[i][j] = '#';

		if (k == word.length() - 1) {
			return true;
		} else if (dfs(board, word, i - 1, j, k + 1) ||
			dfs(board, word, i + 1, j, k + 1) ||
			dfs(board, word, i, j - 1, k + 1) ||
			dfs(board, word, i, j + 1, k + 1)) {
			board[i][j] = temp;
			return true;
		}

	} else {
		return false;
	}

	return false;
}

Method 2 - DFS + Trie

If the current candidate does not exist in all words’ prefix, we can stop backtracking immediately. This can be done by using a trie structure.

Note that here we are passing visited array, but in Method 1, we didn’t pass, as we were modifiying our board[][] matrix by setting # to mark visited.

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public List<String> findWords(char[][] board, String[] words) {
	List<String> ans = new LinkedList<String> ();
	Trie trie = new Trie();
	for (String word: words) {
		trie.insert(word);
	}

	int m = board.length;
	int n = board[0].length;

	boolean[][] visited = new boolean[m][n];

	for (int i = 0; i<m; i++) {
		for (int j = 0; j<n; j++) {
			dfs(board, visited, "", i, j, trie);
		}
	}

	return new ArrayList<String> (result);
}

public void dfs(char[][] board, boolean[][] visited, String str, int i, int j, Trie trie) {
	int m = board.length;
	int n = board[0].length;

	if (i<0 || j<0 || i >= m || j >= n) {
		return;
	}

	if (visited[i][j]) {
		return;
	}

	str = str + board[i][j];

	if (!trie.startsWith(str)) {
		return;
	}
	if (trie.search(str)) {
		result.add(str);
	}

	visited[i][j] = true;
	dfs(board, visited, str, i - 1, j, trie);
	dfs(board, visited, str, i + 1, j, trie);
	dfs(board, visited, str, i, j - 1, trie);
	dfs(board, visited, str, i, j + 1, trie);
	visited[i][j] = false;
}
Trie Class
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//Trie Node
class TrieNode {
	public TrieNode[] children = new TrieNode[26];
	public String item = "";
}

//Trie
class Trie {
	public TrieNode root = new TrieNode();

	public void insert(String word) {
		TrieNode node = root;
		for (char c: word.toCharArray()) {
			if (node.children[c - 'a'] == null) {
				node.children[c - 'a'] = new TrieNode();
			}
			node = node.children[c - 'a'];
		}
		node.item = word;
	}

	public boolean search(String word) {
		TrieNode node = root;
		for (char c: word.toCharArray()) {
			if (node.children[c - 'a'] == null)
				return false;
			node = node.children[c - 'a'];
		}
		if (node.item.equals(word)) {
			return true;
		} else {
			return false;
		}
	}

	public boolean startsWith(String prefix) {
		TrieNode node = root;
		for (char c: prefix.toCharArray()) {
			if (node.children[c - 'a'] == null)
				return false;
			node = node.children[c - 'a'];
		}
		return true;
	}
}