HackerRank Array Pairs problem solution

In this tutorial, we are going to solve or make a solution to the Array Pairs problem. so here we have given an array of n integers and we need to find and print the total number of pairs such that the multiplication of the paired element is less than the max value of the array.

Problem solution in C++ programming.

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

#define rep(i,n) for(int i=0;i<n;i++)
#define ll long long int
#define f first
#define s second
#define pi pair<ll,ll>
#define pii pair<pi,ll>
#define f first
#define s second
#define pb push_back
#define mod 1000000007
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)

int N;
int A[1000011];
int L[1000011];
int R[1000011];
vector<int>g[1000011];
ll bt[1000011];
int maxn;
void update(int ind, int val) {
    while(ind <= maxn) {
        bt[ind] += val;
        ind += (ind & -ind);
    }
}
ll query(int ind) {
    ll ans = 0;
    while(ind > 0) {
        ans += bt[ind];
        ind -= (ind & -ind);
    }
    return ans;
}
vector<int>V;
int find_ind(int x) {
    if(V.back() <= x) return V.size();
    return upper_bound(V.begin(), V.end(), x) - V.begin();
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin >> N;
    set<int>S;
    unordered_map<int, int>M;

    for(int i = 1; i <= N; i++) {
        cin >> A[i];
        assert(A[i] >= 1 and A[i] <= 1000000000);
        S.insert(A[i]);
    }
    vector<pi>window;
    for(int i = 1; i <= N; i++) {
        while(window.size() > 0 and window.back().f < A[i]) window.pop_back();
        if(window.size() == 0) L[i] = 1;
        else {
            L[i] = window.back().s + 1;
        }
        window.pb(mp(A[i], i));
    }
    window.clear();
    for(int i = N; i >= 1; i--) {
        while(window.size() > 0 and window.back().f <= A[i]) window.pop_back();
        if(window.size() == 0) R[i] = N;
        else {
            R[i] = window.back().s - 1;
        }
        window.pb(mp(A[i], i));
    }

    for(int i = 1; i <= N; i++) {
        if(i - L[i] <= R[i] - i) {
            for(int j = L[i]; j < i; j++) {
                g[i - 1].pb(-A[i] / A[j]);
                g[R[i]].pb(A[i] / A[j]);
                //S.insert(A[i]/A[j]);
            }

            g[i].pb(-1);
            g[R[i]].pb(1);
        } else {

            for(int j = i + 1; j <= R[i]; j++) {
                g[L[i] - 1].pb(-A[i] / A[j]);
                g[i].pb(A[i] / A[j]);
                //S.insert(A[i]/A[j]);
            }

            g[L[i] - 1].pb(-1);
            g[i - 1].pb(1);
        }
    }
    maxn = S.size() + 2;
    int cnt = 1;
    for(set<int>::iterator it = S.begin(); it != S.end(); it++) {
        M[*it] = cnt++;
    }
    ll ans = 0;
    int r;
    V = vector<int>(S.begin(), S.end());
    for(int i = 1; i <= N; i++) {
        update(M[A[i]], 1);
        for(int j = 0; j < g[i].size(); j++) {
            r = find_ind(abs(g[i][j]));
            if(g[i][j] < 0) {
                ans -= query(r);
            } else {
                ans += query(r);
            }
        }
    }
    cout << ans;
}