# 5ms Simple DFS Java Solution

• This problem can be dealt with in a very simple way. A 'O' will not be surrounded by all sides only if it is linked (directly or through another 'O') to a 'O' that is on the boundary row or column.

This means that if all 4 boundaries have only 'X' then all the characters can be switched to 'X'

For example if your region is defined like this

XXXX
XOOX
XOOX
XXXX then we can convert all the 'O' to'X's .

Solution: Run through the boundaries and perform dfs when you come across a 'O' . Switch all adjoining 'O' to another character. (I have chosen to switch them to 'Y'). This way all th e'O's that are not surrounded completely by 'X' are switched to 'Y's .

After simply run through the matrix again and switch all 'Y' to 'O' and all 'O' to 'X'.

``````public class Solution {

public void solve(char[][] board) {

if(board==null || board.length==0) return;

int row = board.length;
int col = board[0].length;

//check first and last col
for(int i=0;i<row;i++){

if(board[i][0]=='O') getEmAll(board,i,1);
if(board[i][col-1]=='O') getEmAll(board,i,col-2);
}

//check first and last  row
for(int i=0;i<col;i++){

if(board[0][i]=='O') getEmAll(board,1,i);
if(board[row-1][i]=='O') getEmAll(board,row-2,i);
}

//Switch all 'O's to 'X's and 'Y's to 'O's
for(int i=1;i<row-1;i++)
{
for(int j=1;j<col-1;j++)
if(board[i][j]=='Y') board[i][j]='O';
else if(board[i][j]=='O') board[i][j]='X';
}
}

public void getEmAll(char[][] board, int row, int col){

if(row>=board.length-1 || row<=0 || col>=board[0].length-1 || col<=0) return;

if(board[row][col]=='X' || board[row][col]=='Y') return;
if(board[row][col]=='O') board[row][col]='Y';

getEmAll(board,row+1,col);
getEmAll(board,row,col+1);
getEmAll(board,row-1,col);
getEmAll(board,row,col-1);

}

}
``````

• @junDrone Nice solution! Leaving the boarder unchanged is good enough to avoid StackOverflow.

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