# Word Search

## Question

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example

Given board =

``````[
"ABCE",
"SFCS",
]
``````

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

Tags

## Analysis

1. 考虑board的边界
2. cell是否已经被遍历过（临时转换为特定标识字符，比如'#'，recursion返回后再重置回原来的值，这样可以省去额外开辟二维数组visited[i][j]的空间复杂度)
3. 搜索结束的标志是`k == word.length()`

Note:

``````board[i][j] ^= 256;
``````

``````board[i][j] ^= 256;
``````

## Solution

DFS - with additional space for `visited[][]` matrix (12ms 52.97% AC)

``````class Solution {
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0 || board[0].length == 0) {
return false;
}
if (word.length() == 0) {
return true;
}
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++) {
if (existHelper(board, visited, i, j, word, 0)) return true;
}
}
return false;
}
private boolean existHelper(char[][] board, boolean[][] visited, int row, int col, String word, int index) {
if (index == word.length()) return true;
if (row < 0 || col < 0 || row >= board.length || col >= board[0].length) return false;
if (word.charAt(index) != board[row][col] || visited[row][col]) return false;

visited[row][col] = true;
int[] dx = new int[] {0, 1, 0, -1};
int[] dy = new int[] {1, 0, -1, 0};

for (int i = 0; i < dx.length; i++) {
if ( existHelper(board, visited, row + dx[i], col + dy[i], word, index + 1)) {
return true;
}
}
visited[row][col] = false;
return false;
}

}
``````

DFS - using temporary `'#'` mark visited element (15 ms 36.92% AC)

``````public class Solution {
/**
* @param board: A list of lists of character
* @param word: A string
* @return: A boolean
*/
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0 || board[0].length == 0) {
return false;
}
if (word.length() == 0) {
return true;
}

for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == word.charAt(0)) {
boolean result = dfs(board, word, i, j, 0);
if (result) {
return true;
}
}
}
}
return false;
}

private boolean dfs(char[][] board, String word, int i, int j, int k) {
if (board == null || board.length == 0 || board[0].length == 0) {
return false;
}
if (word.length() == 0) {
return true;
}

if (word.length() == k) {
return true;
}

boolean result = false;
if (insideBoard(board, i, j) && board[i][j] == word.charAt(k)) {
char temp = board[i][j];
board[i][j] = '#';
result = 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 result;
}

private boolean insideBoard(char[][] board, int i, int j) {
return (i >= 0 && i < board.length && j >= 0 && j < board[0].length);
}
}
``````