DSA Tracker

Problem

Pattern this teaches:Graph BFS and DFS

Medium

Rotting Oranges

A medium Graph interview guide. Task statement, worked examples, intuition, and step-by-step solutions.

Topic
Graph
Sheets
3
Core for
12 roles

The problem

Given a grid where each cell is 0 (empty), 1 (fresh orange), or 2 (rotten orange), every minute all fresh oranges adjacent to a rotten orange become rotten. Return the minimum number of minutes until no fresh orange remains. If impossible, return -1.

Example 1

Input
grid = [[2,1,0],[1,1,1],[0,1,2]]
Output
2
Why
Minute 1: oranges at (0,1),(1,0),(1,2),(2,1) rot. Minute 2: orange at (1,1) rots. All fresh oranges rotten in 2 minutes.

Example 2

Input
grid = [[2,0,1],[0,0,0],[1,0,1]]
Output
-1
Why
Rotten orange at (0,0) cannot cross the row of zeros at row 1. Fresh oranges at (0,2), (2,0), (2,2) remain unrotted. Returns -1.

Constraints

  • 1 <= grid.length, grid[i].length <= 10
  • grid[i][j] is 0, 1, or 2

How to think about it

Updated 2026-09-29

Rot spreads outward simultaneously across all fronts, which is the definition of multi-source breadth-first search. Enqueue every initially rotten orange at minute 0 and advance level by level. Count fresh oranges initially: if any fresh oranges remain after the queue empties, complete rotting was impossible.

Approaches, worst first

  1. Step-by-step full matrix scan

    time O((m * n)^2) · space O(m * n)

    Iteratively scan the entire grid each minute to rot fresh oranges adjacent to rotten ones, using temporary markings to prevent newly rotted oranges from spreading within the same minute.

  2. Multi-source BFS with level queueWrite this one

    time O(m * n) · space O(m * n)

    Seed a queue with all cells containing 2 and count total fresh oranges. Process the queue level by level, incrementing minutes only when at least one fresh orange gets contaminated. Return minutes if fresh count reaches zero, else -1.

Where people lose marks · 3
  • If there are initially 0 fresh oranges, the answer is 0, not -1 and not 1 minute.
  • Incrementing the minute counter when popping cells even when no neighbor is infected inflates the time by 1 at the end.
  • Mutating a cell to 2 only upon popping it from the queue allows multiple neighbors to enqueue the same fresh orange multiple times.

The theory behind it

Graph — the ground this problem stands on. All Graph problems

What Graph is

A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.

When to reach for it

Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.

How the pattern works

First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.

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