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In this post, you will find the solution for the **Surrounded Regions** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Surrounded Regions– LeetCode Problem

Surrounded Regions– LeetCode Problem

**Problem:**

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:Surrounded regions should not 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.

**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 Solutions

Surrounded Regions Solution in C++:

class Solution { public: void solve(vector<vector<char>>& board) { if (board.empty()) return; const int m = board.size(); const int n = board[0].size(); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (i * j == 0 || i == m - 1 || j == n - 1) dfs(board, i, j); for (vector<char>& row : board) for (char& c : row) if (c == '*') c = 'O'; else if (c == 'O') c = 'X'; } private: // mark 'O' grids that stretch from four sides with '*' void dfs(vector<vector<char>>& board, int i, int j) { if (i < 0 || i == board.size() || j < 0 || j == board[0].size()) return; if (board[i][j] != 'O') return; board[i][j] = '*'; dfs(board, i + 1, j); dfs(board, i - 1, j); dfs(board, i, j + 1); dfs(board, i, j - 1); } };

Surrounded Regions Solution in Java:

class Solution { public void solve(char[][] board) { if (board.length == 0) return; final int m = board.length; final int n = board[0].length; for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (i * j == 0 || i == m - 1 || j == n - 1) dfs(board, i, j); for (char[] row : board) for (int i = 0; i < row.length; ++i) if (row[i] == '*') row[i] = 'O'; else if (row[i] == 'O') row[i] = 'X'; } // mark 'O' grids that stretch from four sides with '*' private void dfs(char[][] board, int i, int j) { if (i < 0 || i == board.length || j < 0 || j == board[0].length) return; if (board[i][j] != 'O') return; board[i][j] = '*'; dfs(board, i + 1, j); dfs(board, i - 1, j); dfs(board, i, j + 1); dfs(board, i, j - 1); } }

Surrounded Regions Solution in Python:

class Solution: def solve(self, board: List[List[str]]) -> None: if not board: return m = len(board) n = len(board[0]) def dfs(i: int, j: int) -> None: if i < 0 or i == m or j < 0 or j == n: return if board[i][j] != 'O': return board[i][j] = '*' dfs(i + 1, j) dfs(i - 1, j) dfs(i, j + 1) dfs(i, j - 1) for i in range(m): for j in range(n): if i * j == 0 or i == m - 1 or j == n - 1: dfs(i, j) for row in board: for i, c in enumerate(row): row[i] = 'O' if c == '*' else 'X'