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

Rotten Oranges Matrix Problem using BFS

Problem Statement

You are given an m x n grid representing a box of oranges. Each cell of the grid can have one of the following values:

  • 0: an empty cell
  • 1: a fresh orange
  • 2: a rotten orange

Each minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Your task is to return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.

Examples

Grid Output Description
[[2,1,1],[1,1,0],[0,1,1]]
4
All oranges become rotten in 4 minutes
[[2,1,1],[0,1,1],[1,0,1]]
-1
Some fresh oranges are unreachable
[[0,2]]
0
No fresh orange to rot
[[1]]
-1
One fresh orange with no rotten source
[[2,2,2],[2,1,2],[2,2,2]]
1
Fresh orange surrounded by rotten ones
[[2,1,1,0],[0,1,1,1],[1,0,1,0],[0,1,1,2]]
-1
Larger 4x4 grid with reachable fresh oranges
[[2,1,0,2],[1,1,1,1],[0,1,0,1],[2,1,1,2],[2,2,2,2]]
2
5x4 grid with multiple rotten sources

Visualization Player

Solution

Understanding the Problem

We are given a 2D grid (matrix) that represents a box of oranges. Each cell in the grid can have one of the following values:

  • 0 — an empty cell (no orange)
  • 1 — a fresh orange
  • 2 — a rotten orange

Every minute, a rotten orange will cause all directly adjacent fresh oranges (up, down, left, right) to become rotten. Our task is to calculate the minimum number of minutes that must pass until no cell has a fresh orange. If it's not possible for all the fresh oranges to rot, we should return -1.

Step-by-Step Solution with Example

Step 1: Understand the input

Let’s consider the following input grid:

[
  [2,1,1],
  [1,1,0],
  [0,1,1]
]

Here, we have one rotten orange at the top-left corner (0,0), and the rest are fresh (1s), except a few empty cells (0s).

Step 2: Choose the right strategy

To simulate the spread of rot over time, we use Breadth-First Search (BFS). We treat all initially rotten oranges as the starting points of BFS and process them level by level (minute by minute).

Step 3: Initialize the BFS queue

We enqueue all the rotten oranges along with their time = 0. We also count how many fresh oranges are initially present. This helps us track when all fresh ones have been rotted.

Step 4: BFS traversal

For each rotten orange in the queue, we visit its four neighbors (up, down, left, right). If any neighbor is a fresh orange, we turn it rotten, decrement the fresh count, and enqueue it with time +1.

Step 5: Track the time

We maintain a variable time to track the number of minutes taken to rot all reachable fresh oranges. It is updated as we process new levels in the BFS queue.

Step 6: Check if all fresh oranges are rotted

Once the BFS is done, we check if any fresh oranges are left. If so, return -1 because they were unreachable. Otherwise, return the maximum time it took to rot all oranges.

Step 7: Apply to our example

Initially rotten orange at (0,0) rots its neighbors one by one:

  • Minute 1: (0,1), (1,0) become rotten
  • Minute 2: (0,2), (1,1)
  • Minute 3: (2,1)
  • Minute 4: (2,2)

All fresh oranges are now rotten. Total time taken: 4 minutes.

Edge Cases

Case 1: All oranges already rotten or no fresh oranges

If there are no fresh oranges, return 0 since no rotting is needed.

Case 2: Fresh oranges exist but no rotten oranges

There is no way to rot the fresh oranges. Return -1.

Case 3: Some fresh oranges are isolated

If a fresh orange is completely surrounded by empty cells or other fresh ones and cannot be reached by a rotten one, it will remain fresh. Again, return -1.

Finally

This problem helps us understand how to simulate processes over time using BFS in a grid. Always begin by identifying the sources (in this case, initially rotten oranges) and process outward in waves. By tracking time and counting fresh oranges, we can determine the exact moment all fresh oranges have been infected—or if it's impossible.

Algorithm Steps

  1. Initialize a queue and count of fresh oranges.
  2. Add all initially rotten oranges to the queue with timestamp 0.
  3. While the queue is not empty:
    1. Pop an orange from the queue.
    2. For each of the 4 directions:
      1. If the adjacent cell has a fresh orange, rot it, decrement fresh count, and add to queue with incremented time.
  4. If fresh count becomes 0, return the last recorded time. Else return -1.

Code

JavaScript
function orangesRotting(grid) {
  const rows = grid.length, cols = grid[0].length;
  const queue = [];
  let fresh = 0, time = 0;

  for (let r = 0; r < rows; r++) {
    for (let c = 0; c < cols; c++) {
      if (grid[r][c] === 2) queue.push([r, c, 0]);
      if (grid[r][c] === 1) fresh++;
    }
  }

  const directions = [[0,1],[1,0],[0,-1],[-1,0]];

  while (queue.length > 0) {
    const [r, c, t] = queue.shift();
    time = Math.max(time, t);
    for (const [dr, dc] of directions) {
      const nr = r + dr, nc = c + dc;
      if (nr >= 0 && nr < rows && nc >= 0 && nc < cols && grid[nr][nc] === 1) {
        grid[nr][nc] = 2;
        fresh--;
        queue.push([nr, nc, t + 1]);
      }
    }
  }

  return fresh === 0 ? time : -1;
}

console.log("Minutes to rot all oranges:", orangesRotting([[2,1,1],[1,1,0],[0,1,1]])); // 4
console.log("Minutes to rot all oranges:", orangesRotting([[2,1,1],[0,1,1],[1,0,1]])); // -1

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(m * n)Even in the best case, we must scan the entire matrix to identify the initial rotten and fresh oranges.
Average CaseO(m * n)Each cell is visited at most once — either to enqueue it or to rot it through BFS.
Worst CaseO(m * n)In the worst case, BFS will traverse every cell in the matrix to rot all reachable fresh oranges.

Space Complexity

O(m * n)

Explanation: In the worst case, the queue may hold all the oranges in the grid, especially if they're all rotten or become rotten one after another.


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