Array-Sum Operation - c++

Question

I have written this code using vector. Some case has been passed but others show timeout termination error.
The problem statement is:-
You have an identity permutation of N integers as an array initially. An identity permutation of N integers is [1,2,3,...N-1,N]. In this task, you have to perform M operations on the array and report the sum of the elements of the array after each operation.
The ith operation consists of an integer opi.
If the array contains opi, swap the first and last elements in the array.
Else, remove the last element of the array and push opi to the end of the array.
Input Format
The first line contains two space-separated integers N and M.
Then, M lines follow denoting the operations opi.
Constraints :
2<=N,M <= 10^5
1 <= op <= 5*10^5
Output Format
Print M lines, each containing a single integer denoting the answer to each of the M operations.
Sample Input 0
3 2
4
2
Sample Output 0
7
7
Explanation 0
Initially, the array is [1,2,3].
After the 1st operation, the array becomes[1,2,4] as opi = 4, as 4 is not present in the current array, we remove 3 and push 4 to the end of the array and hence, sum=7 .
After 2nd operation the array becomes [4,2,1] as opi = 2, as 2 is present in the current array, we swap 1 and 4 and hence, sum=7.
Here is my code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
long int N,M,op,i,t=0;
vector<long int > g1;
cin>>N>>M;
if(N>=2 && M>=2) {
g1.reserve(N);
for(i = 1;i<=N;i++) {
g1.push_back(i);
}
while(M--) {
cin>>op;
auto it = find(g1.begin(), g1.end(), op);
if(it != (g1.end())) {
t = g1.front();
g1.front() = g1.back();
g1.back() = t;
cout<<accumulate(g1.begin(), g1.end(), 0);
cout<<endl;
}
else {
g1.back() = op;
cout<<accumulate(g1.begin(), g1.end(), 0);
cout<<endl;
}
}
}
return 0;
}
Please Suggest changes.

Answer 1

Looking carefully in question you will find that the operation are made only on the first and last element. So there is no need to involve a whole vector in it much less calculating the sum. we can calculate the whole sum of the elements except first and last by (n+1)(n-2)/2 and then we can manipulate the first and last element in the question. We can also shorten the search by using (1<op<n or op==first element or op == last element).
p.s. I am not sure it will work completely but it certainly is faster

Answer 2

my guess, let take N = 3, op = [4, 2]
N= [1,2,3]
sum = ((N-2) * (N+1)) / 2, it leave first and last element, give the sum of numbers between them.
we need to play with the first and last elements. it's big o(n).
function performOperations(N, op) {
let out = [];
let first = 1, last = N;
let sum = Math.ceil( ((N-2) * (N+1)) / 2);
for(let i =0;i<op.length;i++){
let not_between = !(op[i] >= 2 && op[i] <= N-1);
if( first!= op[i] && last != op[i] && not_between) {
last = op[i];
}else {
let t = first;
first = last;
last = t;
}
out.push(sum + first +last)
}
return out;
}