From cdbe07d263de1727c918e4661261402b0ae790f4 Mon Sep 17 00:00:00 2001
From: AtiaBabaBa <62164180+AtiaBabaBa@users.noreply.github.com>
Date: Sat, 17 Oct 2020 11:42:19 +0700
Subject: [PATCH] Please commit changes, Thank you.

#Hacktoberfest
---
 Binary Search/Cat/Cat.txt | 63 +++++++++++++++++++++++++++++
 Binary Search/Cat/Sol.cpp | 84 +++++++++++++++++++++++++++++++++++++++
 2 files changed, 147 insertions(+)
 create mode 100644 Binary Search/Cat/Cat.txt
 create mode 100644 Binary Search/Cat/Sol.cpp

diff --git a/Binary Search/Cat/Cat.txt b/Binary Search/Cat/Cat.txt
new file mode 100644
index 0000000..efc943a
--- /dev/null
+++ b/Binary Search/Cat/Cat.txt	
@@ -0,0 +1,63 @@
+This problem from TOI13 (Thailand Olympaid of Informatics 2017).
+I love this problem because I love Cats.
+
+Source: https://www.proprog.ml/tasks/toi13_cat/descs/12575
+
+Time Limit 1000 ms
+Memory Limit 512 MB
+
+N cats stay in line. N is always even because we want to pair the cats which have the same height in
+the consecutive position(Each value of height can have only 1 pair). But the cats stay in a random position.
+It is your task to find the size of the pocket to bring the cats and move it to the right position.
+
+For Example:
+N = 6 (6 cats)
+Cat height = 3 3 2 5 5 2
+We can see that the cats that height equals to 3 and 5 already each with the couple but the cats that
+height equals to 2 didn't sit near to each other.
+So, we will use the pocket that has the size of 2 to bring the two-unit-height cat to sit next to each other.
+It'll become 3 3 2 2 5 5 or 3 3 5 5 2 2.
+
+Input:
+1st line contains amount of N while 1<=N<=2x10^6 or 2000000
+Next N lines is hi stands for the height of cat number i while 1<=i<=N and 1<=hi<=2^31
+Output:
+The smallest size of pocket that can rearrange all cat to the right position.
+
+Example input1:
+6
+3
+3
+2
+5
+5
+2
+Example output1:
+2
+(Like an example)
+
+Example input2:
+6
+3
+5
+2
+2
+5
+3
+Example output2:
+3
+(Use pocket size 3 to bring 2 and 3 cats sit next to each other it'll become 5 5 3 3 2 2 for example)
+
+Example input3:
+6
+2
+2
+3
+3
+5
+5
+Example output3:
+0
+(Doesn't need to use the pocket)
+
+
diff --git a/Binary Search/Cat/Sol.cpp b/Binary Search/Cat/Sol.cpp
new file mode 100644
index 0000000..60e3a8c
--- /dev/null
+++ b/Binary Search/Cat/Sol.cpp	
@@ -0,0 +1,84 @@
+#include<bits/stdc++.h>
+using namespace std;
+vector<long long int> v;
+long long int l=0,r=0,mid,big;
+int unpair;
+long long int data[2000009];
+int main()
+{
+	int n;
+	scanf(" %d",&n);
+	long long int a;
+	for(int i=0;i<n;i++)
+	{
+		scanf(" %lld",&a);
+		v.push_back(a);
+		l=min(l,a);
+		r=max(r,a);
+		data[i]=a;
+		if(i%2==1)
+		{
+			//if the cat that stay in odd array has not the same value of the left cat then it is not in pair.
+			if(unpair==0)
+			{
+				if(a!=data[i-1]) 
+				{
+					unpair=1;
+				}
+			}
+		}
+	}
+	//In the case that every cat stay in pair.
+	if(unpair==0)
+	{
+		printf("0");
+		return 0;
+	}
+	sort(data,data+n);
+	r=n-1;
+	mid=(l+r)/2;
+	int broke=0;
+	//B Search for the size of pocket.
+	while(1)
+	{
+		if(l==r) break;
+		for(int i=0;i<n;i++)
+		{
+			//If the cat is bigger than the pocket...
+			if(data[mid]<v[i])
+			{
+				if(big==0)
+				{
+					big=v[i];
+				}
+				//And the next larger cat is not sit next to the cat we have to change the pocket size until we find the best size of pocket.
+				else
+				{
+					if(v[i]!=big)
+					{
+						broke=1;
+						break;
+					}
+					else
+					{
+						big=0;
+					}
+				}
+			}
+		}
+		if(broke==0)
+		{
+			r=mid;
+		}
+		else
+		{
+			l=mid+1;
+		}
+		broke=0;
+		mid=(l+r)/2;
+		big=0;
+	}
+	//Print the smallest size of pocket that can rearrange the cats.
+	printf("%d",data[l]);
+	return 0;
+}