Fundamental Idea

Basically, simulate the operations by storing the integer written on the ball at each position and find the resulting permutation.
Let \(val[i]\) be the variable that represents the integer written on the \(i\)-th ball from the left.

Initially, the \(i\)-th (\(1 \leq i \leq N\)) ball from the left has an integer \(i\) written on it.

So let us initialize \(val\) as follows:

vector<int> val(N+1);
for(int i=1;i<=N;i++)val[i] = i;

The \(Q\) operations can be simulated as follows:

for(int i=0;i<Q;i++){
	for(int j=1;j<=N;j++){
		if(val[j]==x[i]){
			if(j!=N)swap(val[j],val[j+1]);
			else swap(val[j],val[j-1]);
			break;
		}
	}
}

However, the time complexity of this simulation is \(\mathrm{O}(NQ)\), which needs an improvement to fit in the time limit under the Constraint of this problem.

Improvement to \(\mathrm{O}(Q)\)

In the code above, we spend \(\mathrm{O}(N)\) time to find \(j\) such that \(val[j]=x[i]\) in each of \(Q\) iterations, but we can manage the variable \(pos[j]\) to store the position of the ball with the integer \(j\) written on it, so that it can be found in an \(\mathrm{O}(1)\) time, which enables us to solve this problem fast enough.

Sample code

#include <bits/stdc++.h>
using namespace std;

int main() {
    
	int N;
	cin>>N;
	
	vector<int> val(N+1),pos(N+1);
	for(int i=1;i<=N;i++)val[i] = i,pos[i] = i;
	
	int Q;
	cin>>Q;
	
	vector<int> x(Q);
	for(int i=0;i<Q;i++)cin>>x[i];
	
	for(int i=0;i<Q;i++){
		int p0 = pos[x[i]];
		int p1 = p0;
		if(p0!=N)p1++;
		else p1--;
		int v0 = val[p0];
		int v1 = val[p1];
		swap(val[p0],val[p1]);
		swap(pos[v0],pos[v1]);
	}
	
	for(int i=1;i<=N;i++){
		if(i!=1)cout<<' ';
		cout<<val[i];
	}
	cout<<endl;
	
    return 0;
}