Time complexity for finding minimum and maximum value of vector's elements in C++

Question

It's the code covering the above problem:

#include <iostream>
using namespace std;
#include <vector>

int main()
{
    unsigned int n;
    cin >> n;
    int elementsOfVector;
    vector <double> wektor;
    for(int i = 0; i<n; i++) {
        cin >> elementsOfVector;
        wektor.push_back(elementsOfVector);
    }
    double min = wektor[0];
    double max = wektor[1];
    if (min > max) {
        min = wektor[1];
        max = wektor[0];
    }
        for(int i = 2; i<n; i++)    {
            if (max < wektor[i]) {
                max = wektor[i];
            }
            else if (min > wektor[i]) {
                min = wektor[i];
            }
        }
    cout << "Min " << min << " max " << max;
    return 0;
}

According to my analysis: Firstly, we have a for loop to assign all vector's elements with values, we make n-iterations so the time complexity of the action is O(n). Then we have a if statement with condition within it where we compare one value to other but there are always just those two values no matter what n input is so we can assume it's O(1) constant complexity in Big-O notation - not sure if this is correct so I would be grateful If anyone could relate. In the second for loop we make n-2 iterations and the operations inside the for loop are simple arithmetic operations and cost 1 so we can avoid it in big O notation: To sum up n + n = 2n O(2n) so total time complexity is O(n). Am I right?

Answer 1

First part for loop. O(n)
Second part if(min>max) O(1)
Third part for loop O(n)

Total O(n)+O(1)+O(n) = O(n)
The third part since you iterator n-2 times, it's O(n).