Surrounded Regions LeetCode Solution – Queslers

Problem – Surrounded Regions

Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example 1:

Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
Output: [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
Explanation: Notice that an 'O' should not be flipped if:
- It is on the border, or
- It is adjacent to an 'O' that should not be flipped.
The bottom 'O' is on the border, so it is not flipped.
The other three 'O' form a surrounded region, so they are flipped.

Example 2:

Input: board = [["X"]]
Output: [["X"]]

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] is 'X' or 'O'.

Surrounded Regions LeetCode Solution in C++

class Solution {
public:
	void solve(vector<vector<char>>& board) {
        int i,j;
        int row=board.size();
        if(!row)
        	return;
        int col=board[0].size();

		for(i=0;i<row;i++){
			check(board,i,0,row,col);
			if(col>1)
				check(board,i,col-1,row,col);
		}
		for(j=1;j+1<col;j++){
			check(board,0,j,row,col);
			if(row>1)
				check(board,row-1,j,row,col);
		}
		for(i=0;i<row;i++)
			for(j=0;j<col;j++)
				if(board[i][j]=='O')
					board[i][j]='X';
		for(i=0;i<row;i++)
			for(j=0;j<col;j++)
				if(board[i][j]=='1')
					board[i][j]='O';
    }
	void check(vector<vector<char> >&vec,int i,int j,int row,int col){
		if(vec[i][j]=='O'){
			vec[i][j]='1';
			if(i>1)
				check(vec,i-1,j,row,col);
			if(j>1)
				check(vec,i,j-1,row,col);
			if(i+1<row)
				check(vec,i+1,j,row,col);
			if(j+1<col)
				check(vec,i,j+1,row,col);
		}
	}
};

Surrounded Regions LeetCode Solution in Python

def solve(self, board):
    if not any(board): return

    m, n = len(board), len(board[0])
    save = [ij for k in range(m+n) for ij in ((0, k), (m-1, k), (k, 0), (k, n-1))]
    while save:
        i, j = save.pop()
        if 0 <= i < m and 0 <= j < n and board[i][j] == 'O':
            board[i][j] = 'S'
            save += (i, j-1), (i, j+1), (i-1, j), (i+1, j)

    board[:] = [['XO'[c == 'S'] for c in row] for row in board]

Surrounded Regions LeetCode Solution in Java

public void solve(char[][] board) {
	if (board.length == 0 || board[0].length == 0)
		return;
	if (board.length < 2 || board[0].length < 2)
		return;
	int m = board.length, n = board[0].length;
	//Any 'O' connected to a boundary can't be turned to 'X', so ...
	//Start from first and last column, turn 'O' to '*'.
	for (int i = 0; i < m; i++) {
		if (board[i][0] == 'O')
			boundaryDFS(board, i, 0);
		if (board[i][n-1] == 'O')
			boundaryDFS(board, i, n-1);	
	}
	//Start from first and last row, turn '0' to '*'
	for (int j = 0; j < n; j++) {
		if (board[0][j] == 'O')
			boundaryDFS(board, 0, j);
		if (board[m-1][j] == 'O')
			boundaryDFS(board, m-1, j);	
	}
	//post-prcessing, turn 'O' to 'X', '*' back to 'O', keep 'X' intact.
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n; j++) {
			if (board[i][j] == 'O')
				board[i][j] = 'X';
			else if (board[i][j] == '*')
				board[i][j] = 'O';
		}
	}
}
//Use DFS algo to turn internal however boundary-connected 'O' to '*';
private void boundaryDFS(char[][] board, int i, int j) {
	if (i < 0 || i > board.length - 1 || j <0 || j > board[0].length - 1)
		return;
	if (board[i][j] == 'O')
		board[i][j] = '*';
	if (i > 1 && board[i-1][j] == 'O')
		boundaryDFS(board, i-1, j);
	if (i < board.length - 2 && board[i+1][j] == 'O')
		boundaryDFS(board, i+1, j);
	if (j > 1 && board[i][j-1] == 'O')
		boundaryDFS(board, i, j-1);
	if (j < board[i].length - 2 && board[i][j+1] == 'O' )
		boundaryDFS(board, i, j+1);
}

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