GFG POTD: Count Spanning Trees in a Graph| Graph-Spanning Tree (Java) | Day 09

Welcome to today's explanation of the GFG POTD (Problem of the Day)! In this video, I break down the strategies, ideas, and core concepts you need to solve the "Count Spanning Trees in a Graph" problem. Understanding graph theory, Laplacian matrices, and Kirchhoff's Matrix Tree Theorem will massively boost your overall problem-solving skills and help you build the solid foundation needed to tackle complex, math-heavy graph questions next time. If you found this video helpful, please show your gratitude by liking the video, and don't forget to subscribe to the channel for more daily coding content. Happy coding! Problem Name: Count Spanning Trees in a Graph Difficulty: Hard Platform: GeeksforGeeks Question: Determine the total number of distinct spanning trees that can be formed from a connected undirected graph with $n$ vertices and $m$ edges. The Approach (Kirchhoff's Matrix Tree Theorem & Gaussian Elimination) 1. Understand the Core Theorem: Counting every single spanning tree manually or via simple traversal is computationally impossible for complex graphs. Instead, the most elegant way to solve this is using Kirchhoff's Matrix Tree Theorem, which gives us a mathematical shortcut using linear algebra. 2.Build the Laplacian Matrix: First, construct an $N \times N$ matrix. For every edge connecting vertex $U$ and $V$, increase the diagonal values (representing the degree of the vertex) at [U][U] and [V][V] by 1. Simultaneously, set the adjacency values at [U][V] and [V][U] to -1. 3. Reduce the Matrix: According to the theorem, the number of spanning trees is equal to any cofactor of the Laplacian matrix. To get this, we simply delete the first row and the first column (row 0 and col 0) to create a reduced (N-1) * (N-1) matrix. 4. Calculate the Determinant: The determinant of this newly reduced matrix is our exact answer! Since calculating determinants recursively is too slow, we use Gaussian Elimination. By swapping rows and eliminating values below the pivot, we convert the matrix into an upper triangular form. The determinant is then simply the product of its diagonal elements. Complexity: Time Complexity: O(N^3 + M) where N is the number of vertices and M is the number of edges. Building the matrix takes O(M) time, and finding the determinant using Gaussian Elimination takes O(N^3) time. Given the constraint N less than 10 , this runs extremely fast! Space Complexity: O(N^2) to store the 2D arrays for the Laplacian and the reduced matrices. Solution Code: https://github.com/Arnab-Pachal1234/GFG-POTD-SOLUTION #gfg #dsa #graphtheory #spanningtree #kirchhoffstheorem #coding #problemoftheday #java #hardquestion #leetcode #competitiveprogramming #datastructures #math