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.

## C++

`// 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;` `}` |

## Python

`# 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` `])` |

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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