## Word Search

06/28/2016 Array Backtracking

## 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 =

[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]


word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

## Solution

Result: Accepted Time: 8 ms

Here should be some explanations.


int dx={0,0,1,-1};
int dy={1,-1,0,0};
int ptr = 0;
bool dfs(int x,int y,char **board,const int row,const int col, const char * word)
{
if(!word[ptr]) return true;
if(x < 0 || y < 0 || x >= row || y >= col || board[x][y]!=word[ptr])
return false;
ptr++;
board[x][y] |= 128;
bool ret = false;
for(int i = 0; i < 4; i++)
if(ret = dfs(x+dx[i],y+dy[i],board,row,col,word))
break;
board[x][y] ^= 128;
ptr--;
return ret;
}

bool exist(char** board, int row, int col, char* word) {
for(int i = 0; i < row; i++)
for(int j = 0; j < col; j++)
if(board[i][j] == word && dfs(i,j,board,row,col,word))
return true;
return false;
}


Complexity Analytics

• Time Complexity: $O(n^2*k)$
• Space Complexity: $O(1)$