HackerEarth Factors problem solution YASH PAL, 31 July 202415 February 2026 In this HackerEarth Factors problem solution, Ben had two numbers M and N. He factorized both the numbers, that is, expressed M and N as a product of prime numbers. M = (P1A1)*(P2A2 )*(P3A3)*… *(PNAN) N = (Q1B1)*(Q2B2 )*(Q3B3)*… *(QNBN) Here, * represents multiplication.P1, P2 …PN are distinct prime numbers. Q1, Q2 …QN are distinct prime numbers.Unfortunately, Ben has lost both the numbers M and N but still he has arrays A and B. He wants to reterive the numbers M and N but he will not be able to. Your task is to determine the minimum and the maximum number of factors their product (that is, M * N) could have. Your task is to tell Ben the minimum and the maximum number of factors that the product of M and N could have.Since the answer can be large, print modulo 1000000007. HackerEarth Factors problem solution.#include<bits/stdc++.h>using namespace std;#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)#define m 1000000007#define test ll t; cin>>t; while(t--)typedef long long int ll;int main() { FIO; test { ll n,i,ans1=1,ans2=1; cin>>n; ll a[n],b[n]; for(i=0;i<n;i++){ cin>>a[i]; } for(i=0;i<n;i++){ cin>>b[i]; } sort(a,a+n); sort(b,b+n,greater<ll>()); for(i=0;i<n;i++){ ans1*=(a[i]+b[i]+1); ans1%=m; ans2*=(a[i]+1); ans2%=m; ans2*=(b[i]+1); ans2%=m; } cout<<ans1<<" "<<ans2<<endl; } return 0;} Second solution#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 14, M = 1e9 + 7;int n;int main() { ios::sync_with_stdio(0), cin.tie(0); int t; cin >> t; while (t--) { cin >> n; int a[n], b[n]; int mx = 1; for (int i = 0; i < n; ++i) { cin >> a[i]; mx = (ll) mx * (a[i] + 1) % M; } for (int i = 0; i < n; ++i) { cin >> b[i]; mx = (ll) mx * (b[i] + 1) % M; } sort(a, a + n); sort(b, b + n, greater<int>()); int mn = 1; for (int i = 0; i < n; ++i) mn = (ll) mn * (a[i] + b[i] + 1) % M; cout << mn << ' ' << mx << 'n'; }} coding problems solutions HackerEarth HackerEarth