[LintCode] Factorization

A non-negative numbers can be regarded as product of its factors.
Write a function that takes an integer n and return all possible combinations of its factors.

 Notice

• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combination.

 

Example

Given n = 8
return [[2,2,2],[2,4]]
// 8 = 2 x 2 x 2 = 2 x 4.

Given n = 1
return []

Given n = 12
return [[2,6],[2,2,3],[3,4]]

 

By definition, factors should be greater than 1 and less than n itself, which means [1, n] is not a valid factorization of n. 

To avoid duplicated fatorization answers, for a given factorization, if we've already found some factors, all factors that are

yet to be found must be >= its previously found factor. 

 

Output condition: when n becomes 1 after dividing all the factors, we should add the current list as one of the factorization.

However, we also need to avoid adding the factorization of [n, 1]. Since we start the dfs from 2, we know that if the case of 

using n itself as a factor, there should be only 1 element in the current list. Thus, when n == 1, we check if the list size is > 1.

If it is, we have a valid factorization. 

 

Solution 1. 

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        getFactorsDfs(results, new ArrayList<Integer>(), n, 2);
        return results;
    }
    private void getFactorsDfs(List<List<Integer>> results, List<Integer> list, int n, int startFactor) {
        if(n == 1) {
            if(list.size() > 1) {
                results.add(new ArrayList<Integer>(list));
            }
            return;
        }
        for(int i = startFactor; i <= n; i++) {
            if(n % i == 0) {
                list.add(i);
                getFactorsDfs(results, list, n / i, i);
                list.remove(list.size() - 1);
            }
        }
    }
}

Solution 2. Efficient than solution 1. 

Solution 1 works correctly but does a lot of unnecessary work. If we uses the fact that any factorization that starts with a factor that is bigger than 

Math.sqrt(n) is going to be a duplicated answer, we can get a more efficient solution. 

Take n == 12 as an example, sqrt(12) is >= 3 but < 4. we have the following factorizations that start with a factor >= 4:

[4, 3]; [6, 2]; they are duplicates of [3, 4] and [2, 6] respectively. 

This is true because for n = a * b, a <= sqrt(n), b>= sqrt(n), if we start to check n's factor from small to big, then we will check case of a * b prior to case of b * a.

As a result, we can cut the stop condition from i <= n to i <= Math.sqrt(n) without losing any valid factorization.

However, this optimization comes with a cost. When the stop condition is i <= Math.sqrt(n), we eliminated all [n, 1] factorizations, not only for the original input n,

but also for n's factors. For example, if n == 12, then [12, 1] will not be considered. When we recursively calls getFactorsDfs on 12's factor 6, we also eliminated the 

factorization of [6,1] for 6 since i is in range [2, Math.sqrt(6)]; This causes the factorization of [2, 6] is missed. The reason is that we should only enfore the rule 

of "actors should be greater than 1 and less than n itself" on the original input n, not on any of its factors. 

To fix this, the highlighted code is added after [startFactor, sqrt(n)] are all checked. It checks if we can use the current n at the current recursive call level as a valid factor.

For the case of the original input n, the list only contains 1 element of n, so this list will not be added as a valid factorization.

For the case of n's factors f, the list already contains at least 2 elements(the divisor d that meets n % d == f and f itself), this list will be added as a valid factorization.

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        getFactorsDfs(results, new ArrayList<Integer>(), n, 2);
        return results;
    }
    private void getFactorsDfs(List<List<Integer>> results, List<Integer> list, int n, int startFactor) {
        if(n == 1) {
            if(list.size() > 1) {
                results.add(new ArrayList<Integer>(list));
            }
            return;
        }
        for(int i = startFactor; i <= Math.sqrt(n); i++) {
            if(n % i == 0) {
                list.add(i);
                getFactorsDfs(results, list, n / i, i);
                list.remove(list.size() - 1);
            }
        }
21         if(n >= startFactor) {
22             list.add(n);
23             getFactorsDfs(results, list, 1, n);
24             list.remove(list.size() - 1);
25         }
    }
}

 

Related Problems

Combination Sum