"Kefa and Company" task from Tarasov's contest

/*
B. Kefa and Company
time limit per test 2 seconds
memory limit per test 256 megabytes
input: standard input
output: standard output
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.

Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!

Input
The first line of the input contains two space-separated integers, n and d (1 ≤ n ≤ 105, ) — the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.

Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 ≤ mi, si ≤ 109) — the amount of money and the friendship factor, respectively.

Output
Print the maximum total friendship factir that can be reached.
*/

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <iomanip>
#include <cmath>
#include <iterator>

using namespace std;

typedef long long int64;

struct keshasFriend
{
    int64 money;
    int64 friendship;
};

bool operator < (keshasFriend & a, keshasFriend & b)
{
    return a.money < b.money;
}

int main()
{
    int64 friendsQty, minDifference;
    cin >> friendsQty >> minDifference;

    vector <keshasFriend> friends(friendsQty);

    for (int i = 0; i < friendsQty; i++)
    {
        cin >> friends[i]. money >> friends[i].friendship;
    }

    sort(friends.begin(), friends.end());

    int64 minMoney = friends[0].money;
    int64 maxFriendship = friends[0].friendship;
    int64 tempFriendship = friends[0].friendship;

    for (int l = 0, r = 1; r < friendsQty;)
    {
        // With the help of two pointers we don't
        // have to go through array more than 1 time

        if (friends[r].money - minMoney < minDifference)
        {
            // If the right border element is good for us,
            // we increase temporary sum, check if it's
            // greater than our max sum and move right pointer

            tempFriendship += friends[r].friendship;
            r++;

            if (tempFriendship > maxFriendship)
            {
                maxFriendship = tempFriendship;
            }
        }
        else
        {
            // If the right border element is not what we're
            // looking for, then we move the left pointer,
            // and therefore change the value of min money
            // (we know for sure that left border element
            // is the smallest because we sorted array
            // that way) and decrease the value
            // of temporary sum by element that
            // we left behind (l - 1)
            l++;
            minMoney = friends[l].money;
            tempFriendship -= friends[l - 1].friendship;
        }
    }

    cout << maxFriendship << endl;

    return 0;
}