Minimum Cost Walk in Weighted Graph | Thought Process | DSU | Leetcode 3108 | codestorywithMIK
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This is the 45thVideo of our Playlist "GRAPHS : Popular Interview Problems".
In this video we will try to solve a very good problem : Valid Parenthesis String | Leetcode 678
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Minimum Cost Walk in Weighted Graph | Detailed Thought Process | Leetcode 3108 | codestorywithMIK
Company Tags : Will update soon
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Graph/Disjoint%20Set/Minimum%20Cost%20Walk%20in%20Weighted%20Graph.cpp
Leetcode Link : https://leetcode.com/problems/minimum-cost-walk-in-weighted-graph/
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Approach Summary :
The approach implemented in the provided Java code solves a problem involving finding the minimum cost in a graph represented by nodes and edges, and then answering queries based on this minimum cost. Here's a summary of the approach:
1. **Union-Find Data Structure**: The solution employs the union-find data structure to efficiently handle connectivity information between nodes in the graph. It maintains an array `parent`, where each element initially points to itself, representing its own component. The `find` method finds the root of the component to which a node belongs, with path compression optimization for efficiency. The `union` method merges two components by setting one component's parent to the root of the other component.
2. **Minimum Cost Calculation**: The solution iterates through the edges of the graph. For each edge, it determines the root nodes of the connected components using the `find` method. It then updates the minimum cost of the merged component using bitwise AND operation with the existing cost. The `cost` array stores the minimum cost for each component.
3. **Answering Queries**: After calculating the minimum cost for each component, the solution processes the queries. For each query, it determines the root nodes of the nodes involved and compares their root nodes. If the root nodes are the same, it retrieves the minimum cost for that component from the `cost` array. If the nodes are in different components or are identical, appropriate values (either the minimum cost or 0) are added to the result array.
Overall, this approach efficiently handles both the computation of minimum costs and answering queries in the given graph.
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✨ Timelines✨
00:00 - Introduction
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