We can sort \(A_1,\ldots,A_N\) to assume \(A_1<\ldots< A_N\) without changing the answer.

If two adjacent terms differ by two, the answer is between them. Since the Constraints guarantee that the answer is uniquely determined, such a position always occurs once.

The complexity of this problem is \(O(N\log N)\).

Sample code (Python)

N=int(input())
A=sorted(list(map(int,input().split())))
for i in range(N-1):
  if A[i+1]-A[i]==2:
    print(A[i]+1)

Sample code (C++)

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

int main(){
  int n;
  cin >> n;
  vector<int>a(n);
  for(auto&x:a)cin >> x;
  sort(a.begin(),a.end());
  for(int i=0;i<n-1;i++){
    if(a[i+1]-a[i]==2){
      cout << a[i]+1 << endl;
    }
  }
}

There is a trick that enables to solve it in a total of \(O(N)\) time, without sorting at all. Can you figure out how to do so?
(Hint: if you subtract the minimum value of \(A_i\) from the sequence, the range becomes from \(0\) to \(N\).)