# 70. Climbing Stairs
Question link is here.
# Question
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
# Solution
A very easy dfs question.
Start DFS at each coordinate the matches the word[0], and keep a visited map to track each coordinate that was visited earlier.
class Solution {
char[][] board = null;
String word = null;
public boolean exist(char[][] b, String w) {
board = b;
word = w;
boolean res = false;
int[][] visited = new int[b.length][b[0].length];
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < board[0].length; j++) {
if(board[i][j] == word.charAt(0)) {
res = helper(i, j, 0, visited);
if(res == true) {
return res;
}
}
}
}
return res;
}
public boolean helper(int x, int y, int index, int[][] visited) {
if(x < 0 || x >= board.length || y < 0 || y >= board[0].length)
return false;
if(board[x][y] != word.charAt(index)) {
return false;
}
if (visited[x][y] == 1) {
return false;
}
if(index == word.length() - 1)
return true;
visited[x][y] = 1;
if (helper(x+1, y, index+1, visited)
|| helper(x-1, y, index+1, visited)
|| helper(x, y-1, index+1, visited)
|| helper(x, y+1, index+1, visited))
return true;
visited[x][y] = 0;
return false;
}
}