From 53c1e0ef9efcaad271deb75e82fe73d202753102 Mon Sep 17 00:00:00 2001
From: Aakash Panchal <51417248+Aakash-Panchal27@users.noreply.github.com>
Date: Tue, 14 Jan 2020 19:33:56 +0530
Subject: [PATCH] Update DSU.md
---
Akash Articles/DSU.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/Akash Articles/DSU.md b/Akash Articles/DSU.md
index e581dc8..1e8d5b7 100644
--- a/Akash Articles/DSU.md
+++ b/Akash Articles/DSU.md
@@ -151,7 +151,7 @@ So every time we run this function, it will re-connect every vertex on the path
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### Quiz Time
-Can you write the iterative version of the above $FIND(X)$ function with path compression?
+Can you write the iterative version of the above $\text{FIND}(X)$ function with path compression?
Answer:
```
@@ -204,7 +204,7 @@ int Find(x)
$Union(X,Y)$ operation first of all finds root element of both the disjoint sets containing X and Y respectively. Then it connects the root element of one of the disjoint set to the another.
-Well, how do we decide which root will connet to which? If we do it randomly then it may increase the tree height up to O(N), which means that the next $Find(x)$ operation will take O(N) time. Can we do better?
+Well, how do we decide which root will connet to which? If we do it randomly then it may increase the tree height up to $O(N)$, which means that the next $Find(x)$ operation will take $O(N)$ time. Can we do better?
Yes, we have two standard techniques: **By size and By rank**.