LeetCode 130: Surrounded Regions Solution

Master LeetCode problem 130 (Surrounded Regions), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

130. Surrounded Regions

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) starting from the boundary cells (cells with 'O' on the edges of the board). We mark all cells connected to these boundary cells as non-surrounded regions. Then, we iterate through the board and mark any remaining 'O' cells as 'X' as they are surrounded regions.

Time complexity: O(m * n) where m is the number of rows and n is the number of columns in the board.
Space complexity: O(m * n) for the recursive stack space.

Solution Code

class Solution {
    public void solve(char[][] board) {
        if (board == null || board.length == 0) {
            return;
        }
        
        int rows = board.length;
        int cols = board[0].length;
        
        for (int i = 0; i < rows; i++) {
            dfs(board, i, 0);
            dfs(board, i, cols - 1);
        }
        
        for (int j = 0; j < cols; j++) {
            dfs(board, 0, j);
            dfs(board, rows - 1, j);
        }
        
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                } else if (board[i][j] == '1') {
                    board[i][j] = 'O';
                }
            }
        }
    }
    
    private void dfs(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != 'O') {
            return;
        }
        
        board[i][j] = '1';
        dfs(board, i + 1, j);
        dfs(board, i - 1, j);
        dfs(board, i, j + 1);
        dfs(board, i, j - 1);
    }
}

Try It Yourself

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