Find if the word exists in the grid (Java)

Spread the love

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 cells, 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.

Java Solution

This problem can be solved by using a typical DFS algorithm. 


public boolean exist(char[][] board, String word) {
    int m = board.length;
    int n = board[0].length;
 
    boolean result = false;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
           if(dfs(board,word,i,j,0)){
               result = true;
           }
        }
    }
 
    return result;
}
 
public boolean dfs(char[][] board, String word, int i, int j, int k){
    int m = board.length;
    int n = board[0].length;
 
    if(i<0 || j<0 || i>=m || j>=n){
        return false;
    }
 
    if(board[i][j] == word.charAt(k)){
        char temp = board[i][j];
        board[i][j]='#';
        if(k==word.length()-1){
            return true;
        }else if(dfs(board, word, i-1, j, k+1)
        ||dfs(board, word, i+1, j, k+1)
        ||dfs(board, word, i, j-1, k+1)
        ||dfs(board, word, i, j+1, k+1)){
            return true;
        }
        board[i][j]=temp;
    }
 
    return false;
}

Similarly, below is another way of writing this algorithm. 


public boolean exist(char[][] board, String word) {
    for(int i=0; i<board.length; i++){
        for(int j=0; j<board[0].length; j++){
            if(dfs(board, word, i, j, 0)){
                return true;
            }
        } 
    }
 
    return false;
}
 
public boolean dfs(char[][] board, String word, int i, int j, int k){
    if(board[i][j]!=word.charAt(k)){
        return false;
    }
 
    if(k>=word.length()-1){
        return true;
    }
 
    int[] di={-1,0,1,0};
    int[] dj={0,1,0,-1};
 
    char t = board[i][j];
    board[i][j]='#';
 
    for(int m=0; m<4; m++){
        int pi=i+di[m];
        int pj=j+dj[m];
        if(pi>=0&&pi<board.length&&pj>=0&&pj<board[0].length&&board[pi][pj]==word.charAt(k+1)){
            if(dfs(board,word,pi,pj,k+1)){
                return true;
            }
        }
    }
 
    board[i][j]=t;
 
    return false;
}

Tags: ,