**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

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'`

.

```
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);
}
}
};
```

```
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]
```

```
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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