Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Surrounded Regions Graph-Based Matrix Problem

Problem Statement

Given a 2D matrix of size N x M filled with characters 'O' and 'X', the goal is to replace all 'O' cells that are completely surrounded by 'X' on all four sides (up, down, left, right).

'O's that are on the border or connected to a border 'O' should not be changed. Only 'O's that are fully enclosed by 'X' must be replaced.

Examples

Input Matrix Output Matrix Description
[["X", "X", "X", "X"], ["X", "O", "O", "X"], ["X", "X", "O", "X"], ["X", "O", "X", "X"]]
[["X", "X", "X", "X"], ["X", "X", "X", "X"], ["X", "X", "X", "X"], ["X", "O", "X", "X"]]
'O's at (1,1), (1,2), (2,2) are surrounded, so flipped
[["O", "O"], ["O", "O"]]
[["O", "O"], ["O", "O"]]
All 'O's are on border or connected to border, so remain unchanged
[["X"]]
[["X"]]
Single cell matrix, no changes
[["O"]]
[["O"]]
Single 'O' on border, not flipped
[] [] Empty matrix, no operation
[["X", "X", "X", "X"], ["X", "O", "O", "X"], ["O", "X", "X", "X"], ["X", "O", "X", "O"], ["O", "O", "X", "X"]]
[["X", "X", "X", "X"], ["X", "X", "X", "X"], ["O", "X", "X", "X"], ["X", "O", "X", "O"], ["O", "O", "X", "X"]]
Internal 'O's surrounded and flipped, border-connected 'O's remain

Visualization Player

Solution

Understanding the Problem

We are given a 2D matrix filled with characters 'X' and 'O'. The goal is to identify and convert all 'O' regions that are completely surrounded by 'X' into 'X's. An 'O' region is defined as a group of connected 'O's (connected horizontally or vertically).

However, if any 'O' in a region is on the border (top row, bottom row, leftmost column, or rightmost column), or is connected to a border 'O', that region is considered safe and should not be converted.

Step-by-Step Solution with Example

Step 1: Identify and understand the input

Let’s take this input matrix:


X X X X
X O O X
X X O X
X O X X

We need to determine which 'O's are surrounded and which are connected to the border.

Step 2: Mark the border-connected 'O's

Start by checking the border cells. If an 'O' is found on the border, perform a BFS/DFS to mark it and all connected 'O's as safe (we can use a temporary marker like '#').

In our example, matrix[3][1] is an 'O' on the border. It is safe. We mark it as '#'. But this 'O' is not connected to any other 'O', so only this cell gets marked.

Step 3: Traverse the matrix to convert unsafe 'O's

Now scan the entire matrix:

  • If a cell is 'O', it means it is not connected to any border 'O'. It is unsafe and must be converted to 'X'.
  • If a cell is '#', it was marked safe earlier and should be converted back to 'O'.

After this step, our matrix becomes:


X X X X
X X X X
X X X X
X O X X

Step 4: Finalize the matrix

Replace all '#' back to 'O'. Now we have the final matrix with all truly surrounded regions converted and border-connected regions preserved.

Edge Cases

  • Empty matrix: No operation needed.
  • All 'X's: No changes required.
  • All 'O's on border: No 'O' should be changed.
  • Matrix with only 1 row or 1 column: All 'O's are border cells and remain unchanged.

Finally

This problem teaches us how to use graph traversal techniques like DFS or BFS to identify connected components, especially under constraints like "not surrounded." Marking safe regions with a temporary symbol is a common trick in matrix-based problems. Always process border-related constraints first and then use a second pass to finalize the result.

Algorithm Steps

  1. Loop through the border rows and columns of the matrix.
  2. If an 'O' is found on the border, start a DFS/BFS from there to mark all connected 'O's as '#'.
  3. After marking, traverse the entire matrix:
    1. If a cell is 'O', change it to 'X'.
    2. If a cell is '#', revert it to 'O'.

Code

JavaScript
function solve(board) {
  const rows = board.length;
  if (rows === 0) return;
  const cols = board[0].length;

  function dfs(r, c) {
    if (r < 0 || c < 0 || r >= rows || c >= cols || board[r][c] !== 'O') return;
    board[r][c] = '#';
    dfs(r + 1, c);
    dfs(r - 1, c);
    dfs(r, c + 1);
    dfs(r, c - 1);
  }

  for (let i = 0; i < rows; i++) {
    if (board[i][0] === 'O') dfs(i, 0);
    if (board[i][cols - 1] === 'O') dfs(i, cols - 1);
  }

  for (let j = 0; j < cols; j++) {
    if (board[0][j] === 'O') dfs(0, j);
    if (board[rows - 1][j] === 'O') dfs(rows - 1, j);
  }

  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (board[i][j] === 'O') board[i][j] = 'X';
      if (board[i][j] === '#') board[i][j] = 'O';
    }
  }
  return board;
}

const board = [
  ["X", "X", "X", "X"],
  ["X", "O", "O", "X"],
  ["X", "X", "O", "X"],
  ["X", "O", "X", "X"]
];

console.log("Updated Board:", solve(board));

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(m × n)Even if there are no 'O's in the matrix, the algorithm must check every border element and traverse the matrix once.
Average CaseO(m × n)Each cell is visited once during border traversal and once during the final transformation, resulting in linear complexity based on the number of cells.
Worst CaseO(m × n)If the entire matrix is filled with 'O's, DFS/BFS could touch each element multiple times, but the total work remains bounded by O(m × n).

Space Complexity

O(m × n)

Explanation: In the worst case, a DFS/BFS queue or recursion stack can hold all cells in the matrix when marking connected regions.


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