$#doanthevu$

Code C++ :

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
    ll n,tich1,tich2;
    cin>>n;
    ll a[n+5];
    for(ll i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    sort(a+1,a+1+n);
    tich1=a[n]*a[n-1];
    tich2=a[1]*a[2];
    cout<<max(tich1,tich2);
    return 0;
}

Idea:

sorting and searching

steps:

Sort input array in increasing order. 

If all elements are positive, then return the product of the last two numbers. 

Else return a maximum of products of the first two and last two numbers.

if you still struggle, here's the python code:

def maxProduct(arr, n):

    arr.sort()
    num1 = num2 = 0

    sum1 = arr[0] * arr[1]

    sum2 = arr[n - 1] * arr[n - 2]
    if (sum1 > sum2):
        num1 = arr[0]
        num2 = arr[1]
    else:
        num1 = arr[n - 2]
        num2 = arr[n - 1]
    print(num1 * num2)
n = int(input())
arr = [int(i) for i in input().split()]
maxProduct(arr, n) 

c++ version, as requested by question author:

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

using namespace std;

int maxProduct(vector<int>& arr) {
    sort(arr.begin(), arr.end());
    int n = arr.size();
    int num1 = 0, num2 = 0;

    int sum1 = arr[0] * arr[1];
    int sum2 = arr[n - 1] * arr[n - 2];
    if (sum1 > sum2) {
        num1 = arr[0];
        num2 = arr[1];
    } else {
        num1 = arr[n - 2];
        num2 = arr[n - 1];
    }
    return num1 * num2;
}

int main() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for(int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    cout << maxProduct(arr) << endl;
    return 0;
}