In this HackerEarth Array Modification problem solution, You are given an array A of N integers. For every index i of the array A, you need to select an index j such that j != i and the product of A[i] and A[j] is maximum for that i. Finally, you need to print the sum of A[i] * A[j] for all the indices i. Since this sum can be large, you need to print it to the modulo 10^9 + 7.
Note: Please refer to the sample explanation section.


HackerEarth Array Modification problem solution


HackerEarth Array Modification problem solution.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){ a%=m;LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL a[100001];
LL ans[100001];int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
assert(cin >> n);
assert(n >= 1 && n <= 100000);
for(int i = 1; i <= n; i++) {
assert(cin >> a[i]);
assert(a[i] >= -1000000000000000000LL && a[i] <= 1000000000000000000LL);
}
vector<LL> positive, negative;
for(int i = 1; i <= n; i++) {
if(a[i] >= 0) {
positive.push_back(a[i]);
}
else {
negative.push_back(a[i]);
}
}
sort(positive.begin() , positive.end());
sort(negative.begin(), negative.end(), greater<LL>());
int positive_cnt = positive.size();
int negative_cnt = negative.size();
for(int i = 1; i <= n; i++) {
if(a[i] == 0) {
ans[i] = 0;
}
else if(a[i] < 0) {
if(negative.size() == 1) {
ans[i] = (a[i] % M) * (positive[0] % M);
}
else{
if(negative.back() == a[i])
ans[i] = (a[i] % M) * (negative[negative_cnt - 2] % M);
else
ans[i] = (a[i] % M) * (negative.back() % M);
}
}
else {
if(positive.size() == 1) {
ans[i] = (a[i] % M) * (negative[0] % M);
}
else{
if(positive.back() == a[i])
ans[i] = (a[i] % M) * (positive[positive_cnt - 2] % M);
else
ans[i] = (a[i] % M) * (positive.back() % M);
}
}
ans[i] %= M;
if(ans[i] < M)
ans[i] += M;
}
LL final_answer = 0;
for(int i = 1; i <= n; i++) {
final_answer += ans[i];
final_answer %= M;
}
cout << final_answer;
}