LeetCode #130 — MEDIUM

Surrounded Regions

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n matrix board containing letters 'X' and 'O', capture regions that are surrounded:

Connect: A cell is connected to adjacent cells horizontally or vertically.
Region: To form a region connect every 'O' cell.
Surround: A region is surrounded if none of the 'O' cells in that region are on the edge of the board. Such regions are completely enclosed by 'X' cells.

To capture a surrounded region, replace all 'O's with 'X's in-place within the original board. You do not need to return anything.

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:

In the above diagram, the bottom region is not captured because it is on the edge of the board and cannot be surrounded.

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'.

Step 01

Brute Force Baseline

Problem summary: You are given an m x n matrix board containing letters 'X' and 'O', capture regions that are surrounded: Connect: A cell is connected to adjacent cells horizontally or vertically. Region: To form a region connect every 'O' cell. Surround: A region is surrounded if none of the 'O' cells in that region are on the edge of the board. Such regions are completely enclosed by 'X' cells. To capture a surrounded region, replace all 'O's with 'X's in-place within the original board. You do not need to return anything.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Union-Find

Example 1

[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]

Example 2

[["X"]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #130: Surrounded Regions
class Solution {
    private final int[] dirs = {-1, 0, 1, 0, -1};
    private char[][] board;
    private int m;
    private int n;

    public void solve(char[][] board) {
        m = board.length;
        n = board[0].length;
        this.board = board;
        for (int i = 0; i < m; ++i) {
            dfs(i, 0);
            dfs(i, n - 1);
        }
        for (int j = 0; j < n; ++j) {
            dfs(0, j);
            dfs(m - 1, j);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == '.') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }

    private void dfs(int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O') {
            return;
        }
        board[i][j] = '.';
        for (int k = 0; k < 4; ++k) {
            dfs(i + dirs[k], j + dirs[k + 1]);
        }
    }
}

// Accepted solution for LeetCode #130: Surrounded Regions
func solve(board [][]byte) {
	m, n := len(board), len(board[0])
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j int)
	dfs = func(i, j int) {
		if i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O' {
			return
		}
		board[i][j] = '.'
		for k := 0; k < 4; k++ {
			dfs(i+dirs[k], j+dirs[k+1])
		}
	}
	for i := 0; i < m; i++ {
		dfs(i, 0)
		dfs(i, n-1)
	}
	for j := 0; j < n; j++ {
		dfs(0, j)
		dfs(m-1, j)
	}
	for i, row := range board {
		for j, c := range row {
			if c == '.' {
				board[i][j] = 'O'
			} else if c == 'O' {
				board[i][j] = 'X'
			}
		}
	}
}

# Accepted solution for LeetCode #130: Surrounded Regions
class Solution:
    def solve(self, board: List[List[str]]) -> None:
        def dfs(i: int, j: int):
            if not (0 <= i < m and 0 <= j < n and board[i][j] == "O"):
                return
            board[i][j] = "."
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                dfs(i + a, j + b)

        m, n = len(board), len(board[0])
        for i in range(m):
            dfs(i, 0)
            dfs(i, n - 1)
        for j in range(n):
            dfs(0, j)
            dfs(m - 1, j)
        for i in range(m):
            for j in range(n):
                if board[i][j] == ".":
                    board[i][j] = "O"
                elif board[i][j] == "O":
                    board[i][j] = "X"

// Accepted solution for LeetCode #130: Surrounded Regions
impl Solution {
    pub fn solve(board: &mut Vec<Vec<char>>) {
        let m = board.len();
        let n = board[0].len();
        let dirs = vec![-1, 0, 1, 0, -1];

        fn dfs(
            board: &mut Vec<Vec<char>>,
            i: usize,
            j: usize,
            dirs: &Vec<i32>,
            m: usize,
            n: usize,
        ) {
            if i >= 0 && i < m && j >= 0 && j < n && board[i][j] == 'O' {
                board[i][j] = '.';
                for k in 0..4 {
                    dfs(
                        board,
                        ((i as i32) + dirs[k]) as usize,
                        ((j as i32) + dirs[k + 1]) as usize,
                        dirs,
                        m,
                        n,
                    );
                }
            }
        }

        for i in 0..m {
            dfs(board, i, 0, &dirs, m, n);
            dfs(board, i, n - 1, &dirs, m, n);
        }
        for j in 0..n {
            dfs(board, 0, j, &dirs, m, n);
            dfs(board, m - 1, j, &dirs, m, n);
        }

        for i in 0..m {
            for j in 0..n {
                if board[i][j] == '.' {
                    board[i][j] = 'O';
                } else if board[i][j] == 'O' {
                    board[i][j] = 'X';
                }
            }
        }
    }
}

// Accepted solution for LeetCode #130: Surrounded Regions
function solve(board: string[][]): void {
    const m = board.length;
    const n = board[0].length;
    const dirs: number[] = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): void => {
        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] !== 'O') {
            return;
        }
        board[i][j] = '.';
        for (let k = 0; k < 4; ++k) {
            dfs(i + dirs[k], j + dirs[k + 1]);
        }
    };
    for (let i = 0; i < m; ++i) {
        dfs(i, 0);
        dfs(i, n - 1);
    }
    for (let j = 0; j < n; ++j) {
        dfs(0, j);
        dfs(m - 1, j);
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (board[i][j] === '.') {
                board[i][j] = 'O';
            } else if (board[i][j] === 'O') {
                board[i][j] = 'X';
            }
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.