Word Search

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.


Naive Way: First came up with a DFS method. Need a hashset for each path to store visited positions.

 public class Solution {  
   class Node{  
     int x,y;  
     Node(int x, int y){  
       this.x = x;  
       this.y = y;  
     }  
     @Override  
     public boolean equals(Object other){  
       if (!(other instanceof Node))  
         return false;  
       Node n = (Node)other;  
       return this.x==n.x && this.y==n.y;  
     }  
     @Override  
     public int hashCode(){  
       return x << 16 | y;  
     }  
   }  
   public boolean exist(char[][] board, String word) {  
     for(int i = 0;i < board.length;i++)  
       for(int j = 0;j < board[i].length;j++){  
           Node node = new Node(i,j);  
           Stack<Node> stack = new Stack<Node>();  
           Set<Node> set = new HashSet<Node>();  
           stack.add(node);  
           set.add(node);  
           if(dfs(board, node, word, 0, stack, set))  
             return true;  
         }  
     return false;  
   }  
   private boolean dfs(char[][] board, Node node, String word, int index, Stack<Node> stack, Set<Node> set){  
     /* ending cases */  
     // find a path  
     if(index == word.length()) return true;  
     // invalid node  
     if(node.x < 0 || node.x >= board.length || node.y < 0 || node.y >= board[0].length) return false;  
     // not match  
     if(board[node.x][node.y] != word.charAt(index)) return false;  
     // recusive case  
     Node up = new Node(node.x-1, node.y);  
     if(!set.contains(up)){  
       stack.push(up);  
       set.add(up);  
       if(dfs(board, up, word, index+1, stack, set)) return true;  
       stack.pop();  
       set.remove(up);  
     }  
     Node down = new Node(node.x+1, node.y);  
     if(!set.contains(down)){  
       stack.push(down);  
       set.add(down);  
       if(dfs(board, down, word, index+1, stack, set)) return true;  
       stack.pop();  
       set.remove(down);  
     }  
     Node left = new Node(node.x, node.y-1);  
     if(!set.contains(left)){  
       stack.push(left);  
       set.add(left);  
       if(dfs(board, left, word, index+1, stack, set)) return true;  
       stack.pop();  
       set.remove(left);  
     }  
     Node right = new Node(node.x, node.y+1);  
     if(!set.contains(right)){  
       stack.push(right);  
       set.add(right);  
       if(dfs(board, right, word, index+1, stack, set)) return true;  
       stack.pop();  
       set.remove(right);  
     }  
     return false;  
   }  
 }  

And then, I think I am bring too much extra stuff to a simple problem. Better use a 2D array to record used cell.

 public class Solution {  
   public boolean exist(char[][] board, String word) {  
     // edge case  
     if(board == null || board.length == 0) return false;  
     if(word.length() > board.length*board[0].length) return false;  
     boolean[][] used = new boolean[board.length][board[0].length];  
     for(int i = 0;i < board.length;i++){  
       for(int j = 0;j < board[0].length;j++){  
         used[i][j] = true;  
         if(dfs(board, i, j, word, 0, used)) return true;  
         used[i][j] = false;  
       }  
     }  
     return false;  
   }  
   private boolean dfs(char[][] board, int x, int y, String word, int index, boolean[][] used){  
     // ending case  
     if(index==word.length()) return true;  
     if(board[x][y]!=word.charAt(index)) return false;  
     // recursive call  
     if(x-1 >= 0 && !used[x-1][y]){  
       used[x-1][y] = true;  
       if(dfs(board,x-1,y,word,index+1,used)) return true;  
       used[x-1][y] = false;  
     }  
     if(x+1 < board.length && !used[x+1][y]){  
       used[x+1][y] = true;  
       if(dfs(board,x+1,y,word,index+1,used)) return true;  
       used[x+1][y] = false;  
     }  
     if(y-1 >= 0 && !used[x][y-1]){  
       used[x][y-1] = true;  
       if(dfs(board,x,y-1,word,index+1,used)) return true;  
       used[x][y-1] = false;  
     }  
     if(y+1 < board[0].length && !used[x][y+1]){  
       used[x][y+1] = true;  
       if(dfs(board,x,y+1,word,index+1,used)) return true;  
       used[x][y+1] = false;  
     }  
     return false;  
   }  
 }  

but this code get TLE for the longest "aaa.."" string case. And then I saw others' code and found a modified, much more elegant way.

 public class Solution {  
   public boolean exist(char[][] board, String word) {  
     // edge case  
     if(word == null || board == null || board.length == 0) return false;  
     if(word.length() > board.length*board[0].length) return false;  
     boolean[][] used = new boolean[board.length][board[0].length];  
     for(int i = 0;i < board.length;i++)  
       for(int j = 0;j < board[0].length;j++)  
         if(dfs(board, i, j, word, 0, used)) return true;  
     return false;  
   }  
   private boolean dfs(char[][] board, int x, int y, String word, int index, boolean[][] used){  
     // ending case  
     if(index==word.length()) return true;  
     if(x < 0 || x >= board.length || y < 0 || y >= board[0].length) return false;  
     if(used[x][y] || board[x][y]!=word.charAt(index)) return false;  
     // recursive call  
     used[x][y] = true;  
     if(dfs(board,x-1,y,word,index+1,used)) return true;  
     if(dfs(board,x+1,y,word,index+1,used)) return true;  
     if(dfs(board,x,y-1,word,index+1,used)) return true;  
     if(dfs(board,x,y+1,word,index+1,used)) return true;  
     used[x][y] = false;  
     return false;  
   }  
 }