**Array and Queries**

### Problem Statement :

Given an array, you are asked to perform a number of queries and divide the array into what are called, beautiful subsequences. The array A has length n. A function f(A) is defined to be a minimal possible , such that it's possible to divide array A into x beautiful subsequences. Note that each element of an array should belong to exactly one subsequence, and subsequence does not necessarily need to be consecutive. A subsequence S with length len is called beautiful if and only if: len = 1 or Let S' be a sorted version of S. It must hold that for every You need to find modulo . Input Format The first line contains a single integer , representing the length of array . The next line contains the array given as space-separated integers. The next line contains a single integer , representing the number of queries. Each of the lines contain two integers and , which is described above. Output Format Print the required answer in one line. Sample Input 0 5 2 2 1 1 1 2 3 2 5 5 Sample Output 0 11

### Solution :

` ````
Solution in C++ :
special credits to :
Tara Mehta
startreckker@gmail.com
#include <bits/stdc++.h>
using namespace std;
long int mod=1000000007 ;
int main(){
int n;
cin>>n;
multiset<int> set1;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
set1.insert(arr[i]);
}
multiset<int> set2=set1;
int count=0;
while(set2.size()){
count++;
int cur=*(set2.begin());
set2.erase(set2.find(cur));
while((set2.find(cur+1))!=set2.end()){
cur++;
set2.erase(set2.find(cur));
}
}
long long int q;
cin>>q;
long long int ans=0;
for(long long int i=0;i<q;i++){
int id,val;
cin>>id>>val;
id--;
int curr=set1.count(arr[id]);
int prev=set1.count(arr[id]-1);
int next=set1.count(arr[id]+1);
//cout<<'\t'<<curr<<'\t'<<prev<<'\t'<<next<<'\n';
if(prev==0 && next==0)
count--;
else if(curr>max(prev,next))
count--;
else if(curr<=min(prev,next))
count++;
set1.erase(set1.find(arr[id]));
arr[id]=val;
curr=set1.count(arr[id]);
prev=set1.count(arr[id]-1);
next=set1.count(arr[id]+1);
//cout<<'\t'<<curr<<'\t'<<prev<<'\t'<<next<<'\n';
if(prev==0 && next==0)
count++;
else if((curr+1)>max(prev,next))
count++;
else if((curr+1)<=min(prev,next))
count--;
set1.insert(val);
ans=(ans+((i+1)*(count))%mod)%mod;
}
cout<<ans;
return 0;
}
```

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