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.
For example,
Given board =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
Stats
| Frequency | 4 |
| Difficulty | 3 |
| Adjusted Difficulty | 3 |
| Time to use | ---------- |
Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)
Solution
This is a very classical DFS question. Writing this solution fast and precise is very importnt.
The solution is recursive DFS search.
The second code posted below comes from this blog. The code is slightly shorter because it checks visited_array at the beginning of search() method, instead of for each directions. Other than that, it’s basically same solution.
Code
First, my code
public boolean exist(char[][] board, String word) {
if (word.length() == 0) return true;
int m = board.length, n = board[0].length;
for (int i = 0; i < m; i ++) {
for (int j = 0; j < n; j ++) {
if (board[i][j] == word.charAt(0)) {
int[][] visited = new int[m][n];
visited[i][j] = 1;
boolean ans = find(i, j, board, visited, word.substring(1));
if (ans) return true;
}
}
}
return false;
}
private boolean find(int a, int b, char[][] board, int[][] visited,
String word) {
if (word.length() == 0) return true;
int m = board.length, n = board[0].length;
char target = word.charAt(0);
if (a > 0 && visited[a-1][b] == 0 && board[a-1][b] == target) {
visited[a - 1][b] = 1;
boolean ans = find(a - 1, b, board, visited, word.substring(1));
if (ans) return true;
visited[a - 1][b] = 0;
} // top
if (a < m - 1 && visited[a+1][b] == 0 && board[a+1][b] == target) {
visited[a + 1][b] = 1;
boolean ans = find(a + 1, b, board, visited, word.substring(1));
if (ans) return true;
visited[a + 1][b] = 0;
} // bottom
if (b > 0 && visited[a][b-1] == 0 && board[a][b-1] == target) {
visited[a][b - 1] = 1;
boolean ans = find(a, b - 1, board, visited, word.substring(1));
if (ans) return true;
visited[a][b - 1] = 0;
} // left
if (b < n - 1 && visited[a][b+1] == 0 && board[a][b+1] == target) {
visited[a][b + 1] = 1;
boolean ans = find(a, b + 1, board, visited, word.substring(1));
if (ans) return true;
visited[a][b + 1] = 0;
} // right
return false;
}
Second, code from blog
public boolean exist(char[][] board, String word) {
int height = board.length;
int width = board[0].length;
boolean[][] visited = new boolean[height][width];
for (int i = 0; i < height; i++)
for (int j = 0; j < width; j++)
if (search(board, visited, i, j, word, 0))
return true;
return false;
}
private boolean search(char[][] board, boolean[][] visited,
int row, int col, String word, int index) {
if (word.charAt(index) != board[row][col])
return false;
if (index == word.length() - 1)
return true;
int height = board.length;
int width = board[0].length;
visited[row][col] = true;
//up
if (row > 0 && !visited[row - 1][col]
&& search(board, visited, row - 1, col, word, index + 1))
return true;
//down
if (row < height - 1 && !visited[row + 1][col]
&& search(board, visited, row + 1, col, word, index + 1))
return true;
//left
if (col > 0 && !visited[row][col - 1]
&& search(board, visited, row, col - 1, word, index + 1))
return true;
//right
if (col < width - 1 && !visited[row][col + 1]
&& search(board, visited, row, col + 1, word, index + 1))
return true;
// if we did not find the path we need set this position as unvisited
visited[row][col] = false;
return false;
}