second largest element in an array efficient approach with O(n)

#include<iostream>
using namespace std;

int getsecondlargest(int arr[], int n){

int res = -1;
int largest = 0;

for(int i=1; i<n; i++){

if(arr[i]>arr[largest]){
  res = largest;
  largest = i;
}

else if(arr[i]<arr[largest]){
  
  if(res==-1 || arr[i]>arr[res]){
    res = i;
  }
  
}

}

return res;

}