Published Feb 17, 2022
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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.
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false
- 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.
function exist(board: string[][], word: string): boolean {
// it is always good to know the dimension of the 2D board
const ROW: number = board.length;
const COL: number = board[0].length;
// we always need a set for dfs
const visited: Set<string> = new Set<string>();
const dfs = (row: number, col: number, index: number): boolean => {
// recursive call always has base case
// good one
if(index === word.length) return true
// bad one
const key: string = `${row},${col}`;
if(row < 0 || row >= ROW || col < 0 || col >= COL || visited.has(key) || board[row][col] !== word[index]) return false;
// recursive call here
visited.add(key)
let res: boolean = ( dfs(row, col - 1, index + 1) ||
dfs(row - 1, col, index + 1) ||
dfs(row, col + 1, index + 1) ||
dfs(row + 1, col, index + 1))
visited.delete(key)
return res;
}
// driver logic
for(let i = 0; i < ROW; i++){
for(let j = 0; j < COL; j++){
if(dfs(i, j, 0) === true) {
return true
}
}
}
return false;
};