[LeetCode#254] Factor Combinations

Problem:

Numbers can be regarded as product of its factors. For example,

8 = 2 x 2 x 2;
  = 2 x 4.

Write a function that takes an integer n and return all possible combinations of its factors.

Note:

1. Each combination‘s factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
2. You may assume that n is always positive.
3. Factors should be greater than 1 and less than n.

Examples: 
input: 1
output:

[]

input: 37
output:

[]

input: 12
output:

[
  [2, 6],
  [2, 2, 3],
  [3, 4]
]

input: 32
output:

[
  [2, 16],
  [2, 2, 8],
  [2, 2, 2, 4],
  [2, 2, 2, 2, 2],
  [2, 4, 4],
  [4, 8]
]

Analysis:

This problem is not hard, could be easily solved through DFS method. However, I came up with a very complicated method initially.
Initial Idea: each factor must be consisted through multi prime numbers, thus we could dissect the "n" into a collection of prime numbers.
Step 1: compute the collection of prime numbers for n. (which must be unique).
Step 2: use the elements in the collection to generate various combinations.

The above idea is right, but it is too complex. According to problem, as long as the factor is not 1, we could include it. Thus, we not we do this?
Considering the process of dissecting a number:
(((2) * 3 ) * 5 )* 6 = 180

Suppose our target is 180, we can first chop out factor 2. then our target becomes 90.
Then we can continue the process, until our target reachs 1. (which means we fully factorize it). Since this problem asks us to compute all such combinations, we should also try to chop out factor "((2) * 3 )" "(((2) * 3 ) * 5 )". This process could be elegantly achieved through:
<At present, we are in "helper(ret, item, n, i)">
for (int i = start; i <= n; i++) {
    if (n % i == 0) {
        ...
        helper(ret, item, n/i, i);
        ...
    }
}
So elegantly, right?
a. for (int i = start; i <= n; i++), searches through all possible numbers through start to n.
Note: you may ask since "1" is not allowed in the combination, why we allow i <= n. Even n*1 is allowed, but in the recursive process, it no longer the n as the initial one. 2(current n) of 4(inital n), apparently we should allow i to be n, otherwise, we would never reach the base case.
-------------------------------------------------------
if (n == 1) {
    if (item.size() > 1) {
        ret.add(new ArrayList<Integer> (item));
    }
    return;
}
-------------------------------------------------------
Note the case of "n * 1" would never include in to the ret set, since we require "item.size()" must larger than 1.
So elegant and smart? Right!!!!
Also take advantage of item‘s size!!!! Great Programming skill!

b. (n % i == 0), guarantees the current i is a factor. 

c. helper(ret, item, n/i, i), makes us to continue the search with updated target "n/i".
Note: since the same factor could appears repeatedlly, we should start from "i" rather than "i+1".

Solution:

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> ret = new ArrayList<List<Integer>> ();
        helper(ret, new ArrayList<Integer> (), n, 2);
        return ret;
    }

    private void helper(List<List<Integer>> ret, List<Integer> item, int n, int start) {
        if (n == 1) {
            if (item.size() > 1) {
                ret.add(new ArrayList<Integer> (item));
            }
            return;
        }
        for (int i = start; i <= n; i++) {
            if (n % i == 0) {
                item.add(i);
                helper(ret, item, n/i, i);
                item.remove(item.size()-1);
            }
        }
    }
}