Given an m x n grid of characters board and a string word, return trueifwordexists 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?
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
for(int i=0; i<board.size(); i++)
for(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()) // base case (will be reached once entire word is formed || if word is null)
return true;
if(i<0 || j<0 || i>=board.size() || j>=board[0].size() || board[i][j]!=word[0]) // invalid
return false;
char c = board[i][j];
board[i][j] = '*';
string s = word.substr(1); // slice off the already found char
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;
}
};
class Solution {
int[][] dirs = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
int m, n;
public boolean exist(char[][] board, String word) {
this.m = board.length;
this.n = board[0].length;
boolean[][] vis = new boolean[m][n];
boolean ans = false;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] != word.charAt(0))
continue;
vis[i][j] = true;
ans |= solve(board, word, 1, i, j, vis);
vis[i][j] = false;
}
}
return ans;
}
boolean solve(char[][] board, String word, int idx, int x, int y, boolean[][] vis) {
if (idx == word.length())
return true;
boolean ans = false;
for (int[] d : dirs) {
int X = x + d[0], Y = y + d[1];
if (X < 0 || X >= m || Y < 0 || Y >= n || vis[X][Y] || word.charAt(idx) != board[X][Y])
continue;
vis[X][Y] = true;
ans |= solve(board, word, idx + 1, X, Y, vis);
vis[X][Y] = false;
}
return ans;
}
}
class Solution {
int[][] dirs = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
int m, n;
public boolean exist(char[][] board, String word) {
this.m = board.length;
this.n = board[0].length;
boolean[][] vis = new boolean[m][n];
boolean ans = false;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] != word.charAt(0))
continue;
vis[i][j] = true;
ans |= solve(board, word, 1, i, j, vis);
vis[i][j] = false;
}
}
return ans;
}
boolean solve(char[][] board, String word, int idx, int x, int y, boolean[][] vis) {
if (idx == word.length())
return true;
boolean ans = false;
for (int[] d : dirs) {
int X = x + d[0], Y = y + d[1];
if (X < 0 || X >= m || Y < 0 || Y >= n || vis[X][Y] || word.charAt(idx) != board[X][Y])
continue;
vis[X][Y] = true;
ans |= solve(board, word, idx + 1, X, Y, vis);
vis[X][Y] = false;
}
return ans;
}
}
