Merge pull request #399 from eshaanrawat1/main

Added Union Find Data Structure

Data Structures and Algorithms/union_find/README.md

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+# Union Find (Disjoint Set Union) - Implementation and Use
+
+## Table of Contents
+- [Why Union Find?](#why-union-find)
+- [Functions and Examples](#functions-and-examples)
+- [Setup](#setup)
+- [Additional Resources](#additional-resources)
+- [Leetcode Questions](#leetcode-questions)
+
+## Why Union Find?
+Union Find is a popular data structure that allows us to solve many different types of graph
+problems. It works best with undirected graphs, and it allows us to figure out whether a node
+is connected to another node.
+
+Some problems it can be used to solve:
+- Find the minimum spanning tree in a graph (Kruskal's)
+- Check if there is a path between two nodes
+- Finding redundant edges 
+- Representing networks 
+
+
+## Functions and Examples
+Union Find seems complex at first, but it is actually a lot easier when you understand that there are 
+only two functions.
+- Find(n) : returns the parent of a node n
+- Union(n1, n2) : connects n1 and n2 if they are not previously connected
+
+Let's look at an example!  
+```python
+u = UnionFind(7)  # create a UnionFind object with 7 nodes (numbered 0 to 6)
+
+u.union(0, 1)     # connects 0 and 1 together
+u.union(5, 6)     # connects 5 and 6 together
+
+u.find(1)         # returns 0, since 0 is parent of 1
+u.find(5)         # returns 5, since 5 is its own parent
+
+u.union(1, 2)     # connects 2 to the component 0-1
+u.find(2)         # 2s parent is now 0
+
+# Now our structure looks like this
+
+# 0-1-2   3  4  5-6
+
+u.union(1, 6)    # first we find the parents of 1 and 6
+                 # parents are 0, and 5
+                 # connect the smaller component to the bigger
+                 # now 5's parent is 0
+
+u.find(6)       # now this goes:
+                # 6 parent is 5 -> 5 parent is 0 -> 0 is its own parent
+```
+
+And that's it! You can use the sample code to test different examples with Union Find.
+In the code, par keeps track of the parent of each node and rank keeps track of the size of 
+each component.
+
+## Setup
+
+First clone the repo
+ > `cd union_find` to get into this folder.   
+ >  call the verify function anywhere, consider adding ``` if __name__ == '__main__'```  
+ > `python union_find.py` to run the demo
+
+ You can modify the structure in the verify function and play around with it.
+
+ ## Additional Resources
+
+ Here are some resources I found useful when learning: 
+ - Neetcode Graph Videos on YouTube 
+ - William Fiset - Union Find Video on YouTube
+ - Union Find Medium Article by Claire Lee 
+ - Union Find Visualizer - Visualgo 
+
+ ## Leetcode Questions
+ - 200 - Number of Islands
+ - 684 - Redundant Connection
+ - 695 - Max Area of an Island
+ - 827 - Making a Large Island 
+ - 2316 - Count Unreachable Pairs of Nodes in an Undirected Graph
+ - 2421 - Maximum Score of a Good Path
+ - 2709 - Greatest Common Divisor Traversal 
+
+ I hope this was helpful. If there are any mistakes or issues or if you want to contribute to union find, feel free to contact me at rawateshaan0 [at] gmail [dot] com

Data Structures and Algorithms/union_find/union_find.py

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+# Basic implementation of the Union Find data structure
+# Assume we have n nodes labeled from 0 to n - 1
+
+class UnionFind:
+    def __init__(self, n):
+        # every node is originally its own parent
+        self.par = [i for i in range(n)]    
+        # self.par = list(range(n)) -- also valid
+
+        # every node originally is in its own 
+        # component of size 1 - this changes during
+        # the union operation
+        self.rank = [1] * n
+
+    def find(self, n) -> int:
+        '''
+        Finds the parent node of n 
+        '''
+
+        # can be optimized with path compression
+        while n != self.par[n]:
+            n = self.par[n]
+        return n
+
+
+    def union(self, n1, n2) -> bool:
+        '''
+        Connects two nodes together if not 
+        already connected
+        '''
+
+        # find the parent of node 1 and 2
+        p1 = self.find(n1)      
+        p2 = self.find(n2)
+
+        # nodes are already connected
+        # cannot union together
+        if p1 == p2:            
+            return False
+
+        # for efficiency, make bigger component
+        # parent of smaller component - reduces
+        # number of steps we have to take in find()
+
+        if self.rank[p1] >= self.rank[p2]:
+            # p2 is smaller, so when union it has a
+            # new parent, p1
+            self.par[p2] = p1
+
+            # p1 gets all the nodes of p2, increasing
+            # its rank, or size
+            self.rank[p1] += self.rank[p2]
+        else:
+            self.par[p1] = p2
+            self.rank[p2] += self.rank[p1]
+
+        return True
+
+    def nodes_connected(self, n1, n2) -> bool:
+        '''
+        Returns if two nodes are connected
+        '''
+
+        # connected if parent is the same
+        return self.find(n1) == self.find(n2)
+
+
+
+def verify():
+    n = 7
+    u = UnionFind(n)
+
+    # False, nodes not connected
+    print(u.nodes_connected(0, 1))
+
+    # True, just connected 0 and 1
+    u.union(0, 1)
+    print(u.nodes_connected(0, 1))
+
+    # Rank is 2, includes 0 and 1
+    print(u.rank[0])
+
+    u.union(4, 5)
+    u.union(1, 4)
+
+    # True, 0 - 1 and 4 - 5 are connected
+    # 1 to 4 connects both components
+    print(u.nodes_connected(0, 5))