Graph-1: Maximum Number of Fish in a Grid || LeetCode 2658 || BFS Graph Traversal Explained

Tackle the 'Find Maximum Fish in a Grid' Problem with an Efficient BFS Solution! 🌟

In this video, we tackle the "Find Maximum Fish in a Grid" problem, a fun graph-based challenge that uses Breadth-First Search (BFS) to explore connected regions in a grid. This problem is a great exercise to practice connected component analysis and dynamic grid traversal techniques for coding interviews.

What You’ll Learn:
✅ How to traverse a grid using BFS to find connected components with maximum values.
✅ Dynamic tracking of visited cells to optimize grid traversal.
✅ Efficiently calculating sums of connected regions in a 2D grid.
✅ Time and space complexity analysis to understand the algorithm’s efficiency.

Problem Overview:
You are given a 2D grid where each cell contains either a positive integer (representing fish) or 0 (no fish). Your task is to determine the maximum number of fish you can collect from any connected region in the grid. A connected region is defined by adjacent cells (up, down, left, right) with positive values.

📌 Applications:
1️⃣ Geographical Mapping: Identifying regions with maximum resources in geographical data.
2️⃣ Game Development: Implementing logic for collecting resources in grid-based games.
3️⃣ Cluster Analysis: Analyzing and aggregating data in grid-based layouts for statistical insights.

Solution Overview
Step 1: Initialization
Track visited cells with a visited matrix to avoid revisiting.
Initialize a variable max_fish_count to store the maximum sum of connected fish regions.
Step 2: Traverse the Grid
For each cell, start a BFS traversal if the cell contains fish ( greater than 0) and is not yet visited.
Use BFS to explore all connected cells in the region and calculate the total fish count.
Step 3: Update Maximum Count
Update max_fish_count with the maximum value of fish found in any connected region.
Time Complexity:
Grid Traversal:

Every cell is processed once.
Time Complexity: O(M x N), where M is the number of rows and N is the number of columns.
BFS Traversal:

Each cell in a connected region is processed once.
Time Complexity: O(M x N) in the worst case.
Overall Time Complexity:

O(M x N).
Space Complexity:
Visited Matrix:

O(M x N), for storing the visited state of each cell.
BFS Queue:

O(M x N), in the worst case when all cells belong to a single connected region.
Overall Space Complexity:

O(M x N).
👉 Try the Problem Yourself:
https://leetcode.com/problems/maximum-number-of-fish-in-a-grid/description
Why This Tutorial?
The 'Find Maximum Fish in a Grid' problem is an excellent way to practice BFS and understand how to solve connected component problems in a grid. By the end of this tutorial, you’ll have a solid understanding of BFS for real-world applications.

💡 Engage with Us:
Got questions or alternative solutions? Drop your thoughts in the comments! Let’s discuss and learn together.

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Tags:
#GraphAlgorithm, #FindMaximumFish, #BFSAlgorithm, #GridTraversal, #ClusterAnalysis, #GameDevelopment, #GeographicalMapping, #ConnectedComponents, #ResourceOptimization, #BreadthFirstSearch, #2DGrid, #CodingInterview, #Algorithms, #ProgrammingTutorial, #TechInterviews, #PythonProgramming, #CodingChallenge, #Optimization, #DataStructures, #PathFinding, #GridBasedProblems, #DynamicTraversal, #InterviewPreparation, #CodingTips, #ProgrammingSkills, #AlgorithmDesign

Видео Graph-1: Maximum Number of Fish in a Grid || LeetCode 2658 || BFS Graph Traversal Explained канала AlgorithmUz