DSA Visualizations | Hello Interview

Surrounded Regions

DESCRIPTION (credit Leetcode.com)

Given an m x n matrix grid containing only characters 'X' and 'O', modify grid to replace all regions of 'O' that are completey surrounded by 'X' with 'X'.

A region of 'O' is surrounded by 'X' if there is no adjacent path (cells that border each other in the N, W, E, S directions) consisting of only 'O' from anywhere inside that region to the border of the board.

EXAMPLES

Input:


grid = [
["X","X","X","X","O"],
["X","X","O","X","X"],
["X","X","O","X","O"],
["X","O","X","X","X"]
["X","O","X","X","X"]
]

Output:


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

Explanation: The region of O's at grid[1][2] and grid[2][2] is completely surrounded, so we replace them with X's. All the other "O"s are connected to the border by a path of all "O"s, so we leave them as is.

Input:


grid = [
["O","O"],
["O","X"],
]

Output:


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

Explanation: All the "O"s are connected to the border.

Run your code to see results here

Explanation

In order to solve this problem, we can first recognize that there are two types of "Os" in the grid, those that are reachable on a path of connected "O"s starting from the border of the grid, and those that are not.

The "Os" that are reachable from the border are highlighted in light blue, while the ones that aren't a highlighted in dark blue.

The "Os" that are reachable from the border are the ones that we want to keep, while the "Os" that are not reachable from the border are the ones that we want to change to "Xs".

In order for an "O" to be reachable from the border, it must be connected to an "O" that is on the border. This means that we can start from each cell in the border of the grid and use depth-first search to find all the "Os" that are reachable from the border of the grid (by following a path of connected "Os" in the N, E, S, W directions). Whenever we find an "O" that is reachable from the border, we can change its value to "S" to mark it as safe.

All the "Os" that are reachable from the border are marked with "S".

After marking them as safe, we can then iterate through each cell in the grid and change the "Os" that are not marked as safe to "Xs", while also changing the "safe" cells back to "Os".

The final grid.

class Solution:
    def surrounded_regions(self, grid): 
        if not grid:
            return grid
        
        rows = len(grid)
        cols = len(grid[0])

        # recursive function to find all the "O"s that are reachable
        # from the border and mark them as "S"
        def dfs(x, y):
            # return immediately if the cell is out of bounds or is not an "O
            if x < 0 or y < 0 or x >= rows or y >= cols or grid[x][y] != 'O':
                return
            grid[x][y] = 'S'

            # explore the neighboring cells
            dfs(x + 1, y)
            dfs(x - 1, y)
            dfs(x, y + 1)
            dfs(x, y - 1)

        # iniitialize the dfs for the first and last column
        for i in range(rows):
            if grid[i][0] == 'O':
                dfs(i, 0)
            if grid[i][cols - 1] == 'O':
                dfs(i, cols - 1)

         # iniitialize the dfs for the first and last row
        for j in range(cols):
            if grid[0][j] == 'O':
                dfs(0, j)
            if grid[rows - 1][j] == 'O':
                dfs(rows - 1, j)

        # change the "O"s that are not marked as "S" to "X"s and the "S"s back to "O"s
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 'O':
                    grid[i][j] = 'X'
                elif grid[i][j] == 'S':
                    grid[i][j] = 'O'
                    
        return grid

Keeping Track of Visited Cells

In the above solution, marking each cell as "S" also serves to keep track of the visited cells. We only make recursive calls to the neighboring cells if the current cell is an "O". This way, we avoid visiting the same cell multiple times, as a cell that was previously "O" but is now "S" will not be visited again.

Complexity Analysis

The time complexity for this approach is O(M * N), where M is the number of rows and N is the number of columns in the grid. This is because we are iterating through each cell in the grid once.

The space complexity is O(M * N), since we have to allocate space for each recursive call in the call stack, which can be at most O(M * N).

Next: Pacific Atlantic Water Flow

Your account is free and you can post anonymously if you choose.

Loading comments...

Back to Home

Eliminating Pairs

Overview

Container With Most Water

3-Sum

Triangle Numbers

Pointers as Regions

Move Zeroes

Sort Colors

Trapping Rain Water

Variable-Length

Overview

Longest Substring Without Repeating Characters

Longest Repeating Character Replacement

Fixed-Length

Overview

Maximum Sum of Subarrays of Size K

Max Points You Can Obtain From Cards

Max Sum of Distinct Subarrays Length k

Overview

Can Attend Meetings

Insert Interval

Non-Overlapping Intervals

Merge Intervals

Employee Free Time

Overview

Valid Parentheses

Decode String

Longest Valid Parentheses

Monotonic Stack

Overview

Daily Temperatures

Largest Rectangle in Histogram

Overview

Linked List Cycle

Palindrome Linked List

Remove Nth Node From End of List

Reorder List

Swap Nodes in Pairs

Overview

Apple Harvest (Koko Eating Bananas)

Search in Rotated Sorted Array

Overview

Kth Largest Element in an Array

K Closest Points to Origin

Find K Closest Elements

Merge K Sorted Lists

Introduction

Binary Trees

Overview

Level Order Sum

Rightmost Node

Zigzag Level Order

Maximum Width of Binary Tree

Graphs

Overview

Minimum Knight Moves

Rotting Oranges

01-Matrix

Bus Routes

Overview

Word Search

Solution Space Trees

Subsets

Generate Parentheses

Combination Sum

Topological Sort

Overview

Course Schedule

Course Schedule II

Fundamentals

Solving a Question with Dynamic Programming

Counting Bits

Decode Ways

Maximal Square

Unique Paths

Longest Increasing Subsequence

Word Break

Maximum Profit in Job Scheduling

Overview

Best Time to Buy and Sell Stock

Gas Station

Jump Game

Overview

Implement Trie Methods

Prefix Matching

Overview

Count Vowels in Substrings

Subarray Sum Equals K

Spiral Matrix

Rotate Image

Set Matrix Zeroes

Number of Islands

Pacific Atlantic Water Flow

Surrounded Regions

DESCRIPTION (credit Leetcode.com)

Given an m x n matrix grid containing only characters 'X' and 'O', modify grid to replace all regions of 'O' that are completey surrounded by 'X' with 'X'.

EXAMPLES

Input:


grid = [
["X","X","X","X","O"],
["X","X","O","X","X"],
["X","X","O","X","O"],
["X","O","X","X","X"]
["X","O","X","X","X"]
]

Output:


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

Input:


grid = [
["O","O"],
["O","X"],
]

Output:


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

Explanation: All the "O"s are connected to the border.

Run your code to see results here

Explanation

The "Os" that are reachable from the border are highlighted in light blue, while the ones that aren't a highlighted in dark blue.

The "Os" that are reachable from the border are the ones that we want to keep, while the "Os" that are not reachable from the border are the ones that we want to change to "Xs".

All the "Os" that are reachable from the border are marked with "S".

After marking them as safe, we can then iterate through each cell in the grid and change the "Os" that are not marked as safe to "Xs", while also changing the "safe" cells back to "Os".

The final grid.

class Solution:
    def surrounded_regions(self, grid): 
        if not grid:
            return grid
        
        rows = len(grid)
        cols = len(grid[0])

        # recursive function to find all the "O"s that are reachable
        # from the border and mark them as "S"
        def dfs(x, y):
            # return immediately if the cell is out of bounds or is not an "O
            if x < 0 or y < 0 or x >= rows or y >= cols or grid[x][y] != 'O':
                return
            grid[x][y] = 'S'

            # explore the neighboring cells
            dfs(x + 1, y)
            dfs(x - 1, y)
            dfs(x, y + 1)
            dfs(x, y - 1)

        # iniitialize the dfs for the first and last column
        for i in range(rows):
            if grid[i][0] == 'O':
                dfs(i, 0)
            if grid[i][cols - 1] == 'O':
                dfs(i, cols - 1)

         # iniitialize the dfs for the first and last row
        for j in range(cols):
            if grid[0][j] == 'O':
                dfs(0, j)
            if grid[rows - 1][j] == 'O':
                dfs(rows - 1, j)

        # change the "O"s that are not marked as "S" to "X"s and the "S"s back to "O"s
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 'O':
                    grid[i][j] = 'X'
                elif grid[i][j] == 'S':
                    grid[i][j] = 'O'
                    
        return grid

Keeping Track of Visited Cells

Complexity Analysis

The time complexity for this approach is O(M * N), where M is the number of rows and N is the number of columns in the grid. This is because we are iterating through each cell in the grid once.

The space complexity is O(M * N), since we have to allocate space for each recursive call in the call stack, which can be at most O(M * N).

Next: Pacific Atlantic Water Flow

Your account is free and you can post anonymously if you choose.

Loading comments...

BlogBehavioral Interview Examples Interviewing for Experienced SWEs All Blog Posts

Compare UsCompare to Interviewing.io

LinksFAQ Schedule Mock Interviews Pricing Become a Coach Our Coaches Learn System Design Learn DSA

LegalTerms and Conditions Privacy Policy

Contactsupport@hellointerview.com7511 Greenwood Ave North
Unit #4238 Seattle, WA 98103