Maximum factors formed by two numbers

Given a list of N numbers, Find two numbers such that the product of the two numbers has the maximum number of factors. Formally given n numbers a0, a1, a2, …..an. We are required to find two numbers ai and aj such that their multiplication gives the maximum number of factors and return number of factors of ai * aj
Constraints
1 <= N <=100
1 <= ai <= 10^8

Examples:

Input : [4, 3, 8] 
Output : 8
3*4 = 12 which has 6 factors
3*8 = 24 which has 8 factors
4*8 = 32 which has 6 factors
for ai, aj = {3, 8} we have the 
maximum number of factors 8 

One simple approach seems to be to take two numbers with maximum prime factors. This approach might not work as the picked two numbers may have many common factors and their product might not have maximum factors. For example, in [4, 3, 8], (4, 8) is not answer, but 3, 8 is answer.



We consider every pair of numbers. For every pair, we find union of factors and finally return the pair which has maximum count in union.

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// C++ program to find the pair whose 
// product has maximum factors.
#include <bits/stdc++.h>
using namespace std;
  
typedef unordered_map<int,int> u_map;
  
// Returns a map that has counts of individual
// factors in v[]
u_map countMap(vector<int> v)
{
    u_map map;
    for (size_t i = 0; i < v.size(); i++) 
        if (map.find(v[i]) != map.end()) 
            map[v[i]]++;        
        else 
            map.insert({v[i], 1});        
    return map;
}
  
// Given two Numbers in the form of their 
// prime factorized maps. Returns total 
// number of factors of the product of 
// those two numbers
int totalFactors(u_map m1, u_map m2)
{
    // Find union of all factors.
    for (auto it = m2.begin(); it != m2.end(); ++it) {
        if (m1.find(it->first) != m1.end()) 
            m1[it->first] += it->second;        
        else 
            m1.insert({ it->first, it->second });        
    }
  
    int product = 1;
    for (auto it = m1.begin(); it != m1.end(); it++) 
        product *= (it->second + 1);
      
    return product;
}
  
// Prime factorization of a number is represented 
// as an Unordered map
u_map primeFactorize(int n)
{
    vector<int> pfac;
    int temp = 2;
    while (temp * temp <= n) {
        if (n % temp == 0) {
            pfac.push_back(temp);
            n = n / temp;
        }
        else {
            temp++;
        }
    }
    if (n > 1) 
        pfac.push_back(n);
      
    return countMap(pfac);
}
  
int maxProduct(int arr[], int n)
{
    // vector containing of prime factorizations
    // of every number in the array. Every element
    // of vector contains factors and their counts.
    vector<u_map> vum;
    for (int i = 0; i < n; i++) 
        vum.push_back(primeFactorize(arr[i]));
      
    // Consider every pair and find the pair with
    // maximum factors. 
    int maxp = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i == j)
                continue;
            maxp = max(maxp,
                  totalFactors(vum[i], vum[j]));
        }
    }
    return maxp;
}
  
// Driver code
int main()
{
    int arr[] = { 4, 3, 8 };
    cout << maxProduct(arr, 3);
    return 0;
}
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# Python program to find the pair whose 
# product has maximum factors.
from collections import Counter
  
# Returns list of factors of n.
def prime_factorize(n):
    primfac = []
    d = 2
    while d * d <= n:
        while (n % d) == 0:
            primfac.append(d)  
            n //= d
        d += 1
    if n > 1:
       primfac.append(n)
    return Counter(primfac)
  
# Returns total factors in c1 and c2
def total_factors(c1, c2):
        
    c = c1 + c2
    v = c.values()
  
    # calc product of each element + 1
    p = 1
    for i in v:
        p *=(i + 1)
    return p
  
def max_product(arr):
    n = len(arr)
  
    # Loop through all the nc2 possibilities
    pfac = []
    for i in arr:
        pfac.append(prime_factorize(i))
    maxp = 0
    for i, v1 in enumerate(arr):
        for j, v2 in enumerate(arr):
            if i == j:
                continue
            p = total_factors(pfac[i], pfac[j])
            if(p>maxp):
                maxp = p
    return maxp
  
if __name__ == '__main__':
    print max_product([4, 8, 3])
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Output:
8

An alternate approach is to find LCM of all pairs and find factors in all LCMs. Finally return the pair whose LCM has maximum factors.




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