130. Surrounded Regions

Given a 2D board containing ‘X‘ and ‘O‘ (the letter O), capture all regions surrounded by ‘X‘.
A region is captured by flipping all ‘O‘s into ‘X‘s in that surrounded region.
Example:
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X
Explanation:
Surrounded regions shouldn’t be on the border, which means that any ‘O‘ on the border of the board are not flipped to ‘X‘. Any ‘O‘ that is not on the border and it is not connected to an ‘O‘ on the border will be flipped to ‘X‘. Two cells are connected if they are adjacent cells connected horizontally or vertically.






class Solution {
    private static final int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    public void solve(char[][] board) {
        // sanity check 
        if(board == null || board.length == 0 || board[0].length == 0) return;
        // the Os on the boarder will still stay O . do bfs and store them in a quueue, store their coordinates 
        // base case, when neither dir is O anymore 
        // the Os which are not in the queue can be converted to X 
        boolean[][] visited = new boolean[board.length][board[0].length];
        Queue<int[]> queue = new LinkedList<>();
        // traverse the 4 borders and add the O‘s coordinates on the boarder into the queue 
        // up : row : 0, col from 0 to board[0].length
        for(int j = 0; j < board[0].length; j++){
            if(board[0][j] == ‘O‘){
                queue.offer(new int[]{0, j});
                visited[0][j] = true;
            } 
        }
        // down : row: board.length -1, col from 0, to board[0].length
        for(int j = 0; j < board[0].length; j++){
            if(board[board.length - 1][j] == ‘O‘){
                queue.offer(new int[]{board.length -1, j});
                visited[board.length - 1][j] = true;
            } 
        }
        // left : row from 0 to board.length , col is 0
        for(int i = 0; i < board.length; i++){
            if(board[i][0] == ‘O‘){
                queue.offer(new int[]{i, 0});
                visited[i][0] = true;
            } 
        }
        // right : row from 0 to board.length , col is board[0].length
        for(int i = 0; i < board.length; i++){
            if(board[i][board[0].length - 1] == ‘O‘) {
                queue.offer(new int[]{i, board[0].length - 1});
                visited[i][board[0].length - 1] = true;
            }
        }
        
        
        // bfs from boarder, convert O to E , 
        bfs(queue, board, visited);
        
        // after bfs, convert O to X, and convert E to O
        convert(board);    
    }
    private void bfs(Queue<int[]> queue, char[][] board, boolean[][] visited){
        while(!queue.isEmpty()){
            int[] cur = queue.poll();
            int row = cur[0];
            int col = cur[1];
            board[row][col] = ‘E‘;
            for(int[] dir : dirs){
                int newRow = row + dir[0];
                int newCol = col + dir[1];
                // check boundary and if its O, and if visited alredy 
                if(newRow < 0 || newCol < 0 || newRow >= board.length || newCol >= board[0].length || board[newRow][newCol] != ‘O‘ ||visited[newRow][newCol]) continue;
                queue.offer(new int[]{newRow, newCol});
            }
            
        }
        
    }
    
    private void convert(char[][] board){
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(board[i][j] == ‘O‘) board[i][j] = ‘X‘;
                if(board[i][j] == ‘E‘) board[i][j] = ‘O‘;
            }
        }
    }
}