Merge pull request #2009 from laxmikandivalasa/algo3

Added rotten oranges

docs/DSA-Problem-Solution/rotten-oranges.md

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+---
+id: rotten-oranges
+title: "Rotten Oranges Algorithm"
+sidebar_label: "RottenOranges algorithm"
+tags: [Leetcode, BFS, Graph, DSA, Rotten Oranges]
+description: "Solve the Rotten Oranges problem using Breadth-First Search (BFS) to determine the minimum time required for all fresh oranges to rot."
+---
+
+The Rotten Oranges problem is a grid-based problem that involves determining the minimum time required for all fresh oranges to rot given an initial configuration of fresh and rotten oranges.
+
+<Ads />
+
+## Problem Definition
+
+**Given:** A 2D grid where each cell can have one of three values:
+0: an empty cell
+1: a fresh orange
+2: a rotten orange
+
+**Objective:** Return the minimum number of minutes needed for all fresh oranges to become rotten. If all oranges can’t rot, return -1.
+Algorithm Overview
+
+**Breadth-First Search (BFS) Approach:**
+
+Use BFS to simulate the spread of rotting from each rotten orange to adjacent fresh oranges. Each level of BFS represents one minute.
+Initialization:
+
+Initialize a queue with all initial rotten oranges and count the fresh oranges.
+Track the minutes taken for all oranges to rot.
+Processing BFS Levels:
+
+For each rotten orange, attempt to rot adjacent fresh oranges (up, down, left, right).
+Add newly rotten oranges to the queue and decrease the count of fresh oranges.
+Result Evaluation:
+
+<Ads />
+
+If there are no remaining fresh oranges after BFS, return the minutes taken. If fresh oranges remain, return -1.
+Time Complexity
+Time Complexity: O(n * m), where n is the number of rows and m is the number of columns, as each cell is processed at most once.
+Space Complexity: O(n * m) for the BFS queue.
+C++ Implementation
+```
+cpp
+Copy code
+#include <vector>
+#include <queue>
+using namespace std;
+
+int orangesRotting(vector<vector<int>>& grid) {
+    int rows = grid.size();
+    int cols = grid[0].size();
+    queue<pair<int, int>> rottenQueue;
+    int freshOranges = 0;
+    int minutes = 0;
+
+    // Initialize queue with all initially rotten oranges and count fresh oranges
+    for (int i = 0; i < rows; i++) {
+        for (int j = 0; j < cols; j++) {
+            if (grid[i][j] == 2) {
+                rottenQueue.push({i, j});
+            } else if (grid[i][j] == 1) {
+                freshOranges++;
+            }
+        }
+    }
+
+    // Directions for adjacent cells (up, down, left, right)
+    vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
+
+    // Perform BFS to rot adjacent oranges
+    while (!rottenQueue.empty() && freshOranges > 0) {
+        int size = rottenQueue.size();
+        minutes++;
+
+        for (int i = 0; i < size; i++) {
+            auto [x, y] = rottenQueue.front();
+            rottenQueue.pop();
+
+            for (auto [dx, dy] : directions) {
+                int newX = x + dx;
+                int newY = y + dy;
+
+                // Check if the adjacent cell is within bounds and has a fresh orange
+                if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && grid[newX][newY] == 1) {
+                    // Rot the fresh orange
+                    grid[newX][newY] = 2;
+                    freshOranges--;
+                    rottenQueue.push({newX, newY});
+                }
+            }
+        }
+    }
+
+    // If there are fresh oranges left, return -1, otherwise return minutes taken
+    return freshOranges == 0 ? minutes : -1;
+}
+```
+This algorithm efficiently uses BFS to simulate the spread of rot across the grid, ensuring that all reachable fresh oranges rot in the minimum time or determining that it's impossible.