Binge watching

input/code.cpp: In function 'int main(int, char**)':
input/code.cpp:74:14: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<std::pair<int, int> >::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   74 |     while (i < movies.size()) {
      |            ~~^~~~~~~~~~~~~~~

#include <algorithm>
#include <iostream>
#include <utility>
#include <vector>

using namespace std;

template <typename T>
std::ostream &operator<<(std::ostream &os, const vector<T> &obj) {
    bool first = true;
    for (const auto &o : obj) {
        if (first) {
            cout << o;
            first = false;
        } else {
            cout << ' ' << o;
        }
    }
    cout << endl;
    return os;
}
std::ostream &operator<<(std::ostream &os, const vector<pair<int, int>> &obj) {
    for (const auto &o : obj) {
        cout << o.first << ' ' << o.second << endl;
    }
    cout << endl;
    return os;
}

/**
Maija has taken a day off from work and is going to binge watch as many movies
as she can. She has listed out all the movies airing on TV today. Given the
list, find the maximum number of movies she can watch from beginning to end.
Input
The first input line contains an integer n which is the number of movies.
After this, there are n lines that describe the movies. Each line has two
integers a and b which are the starting and ending times of a movie.

Output
Print one integer which is the maximum number of movies.
Constraints

1 \le n \le 2 \cdot 10^5
1 \le a < b \le 10^9

Example
Input:
3
3 5
4 9
5 8

Output:
2
 */

vector<pair<int, int>> movies;

int main(int argc, char **argv) {
    int n;
    cin >> n;
    movies = vector<pair<int, int>>(n);
    for (int i = 0; i < n; ++i) {
        cin >> movies[i].first;
        cin >> movies[i].second;
    }

    sort(movies.begin(), movies.end(),
         [](auto a, auto b) { return a.second < b.second; });
    // cout << movies;
    int i = 0;
    int count = 0;
    int end_time = -1;
    while (i < movies.size()) {
        auto &m = movies[i];
        if (m.first < end_time) {
        } else {
            end_time = m.second;
            count++;
        }
        i++;
    }
    cout << count << endl;
}

Test 1

Verdict: ACCEPTED

Test 2

Verdict: ACCEPTED

Test 3

Verdict: ACCEPTED

Test 4

Verdict: ACCEPTED

Test 5

Verdict: ACCEPTED

Test 6

Verdict: ACCEPTED

Test 7

Verdict: ACCEPTED

Test 8

Verdict: ACCEPTED

Test 9

Verdict: ACCEPTED