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# Word Search LeetCode Solution

## Problem – Word Search

Given an `m x n` grid of characters `board` and a string `word`, return `true` if `word` exists in the grid.

The word can 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.

Example 1:

``````Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true``````

Example 2:

``````Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true``````

Example 3:

``````Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false``````

Constraints:

• `m == board.length`
• `n = board[i].length`
• `1 <= m, n <= 6`
• `1 <= word.length <= 15`
• `board` and `word` consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger `board`?

### Word Search LeetCode Solution in Java

``````public boolean exist(char[][] board, String word) {
char[] w = word.toCharArray();
for (int y=0; y<board.length; y++) {
for (int x=0; x<board[y].length; x++) {
if (exist(board, y, x, w, 0)) return true;
}
}
return false;
}

private boolean exist(char[][] board, int y, int x, char[] word, int i) {
if (i == word.length) return true;
if (y<0 || x<0 || y == board.length || x == board[y].length) return false;
if (board[y][x] != word[i]) return false;
board[y][x] ^= 256;
boolean exist = exist(board, y, x+1, word, i+1)
|| exist(board, y, x-1, word, i+1)
|| exist(board, y+1, x, word, i+1)
|| exist(board, y-1, x, word, i+1);
board[y][x] ^= 256;
return exist;
}
``````

### Word Search LeetCode Solution in Python

``````def exist(self, board, word):
if not board:
return False
for i in xrange(len(board)):
for j in xrange(len(board[0])):
if self.dfs(board, i, j, word):
return True
return False

# check whether can find word, start at (i,j) position
def dfs(self, board, i, j, word):
if len(word) == 0: # all the characters are checked
return True
if i<0 or i>=len(board) or j<0 or j>=len(board[0]) or word[0]!=board[i][j]:
return False
tmp = board[i][j]  # first character is found, check the remaining part
board[i][j] = "#"  # avoid visit agian
# check whether can find "word" along one direction
res = self.dfs(board, i+1, j, word[1:]) or self.dfs(board, i-1, j, word[1:]) \
or self.dfs(board, i, j+1, word[1:]) or self.dfs(board, i, j-1, word[1:])
board[i][j] = tmp
return res
``````

### Word Search LeetCode Solution in C++

``````bool exist(vector<vector<char>>& board, string word) {
for (unsigned int i = 0; i < board.size(); i++)
for (unsigned int j = 0; j < board[0].size(); j++)
if (dfs(board, i, j, word))
return true;
return false;
}

bool dfs(vector<vector<char>>& board, int i, int j, string& word) {
if (!word.size())
return true;
if (i<0 || i>=board.size() || j<0 || j>=board[0].size() || board[i][j] != word[0])
return false;
char c = board[i][j];
board[i][j] = '*';
string s = word.substr(1);
bool ret = dfs(board, i-1, j, s) || dfs(board, i+1, j, s) || dfs(board, i, j-1, s) || dfs(board, i, j+1, s);
board[i][j] = c;
return ret;
}
``````
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##### Conclusion:

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