Buttercola: Leetcode: Factor Combinations

Tuesday, August 25, 2015

Leetcode: Factor Combinations

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:

Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
You may assume that n is always positive.
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]
]

Code (Java):

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (n <= 3) {
            return result;
        }
         
        List<Integer> curr = new ArrayList<Integer>();
         
        getFactorsHelper(n, n, 2, curr, result);
         
        return result;
    }
     
    private void getFactorsHelper(int ori, int n, int start, List<Integer> curr, List<List<Integer>> result) {
        if (n <= 1) {
            result.add(new ArrayList<Integer>(curr));
            return;
        }
         
        for (int i = start; i <= n && i < ori; i++) {
            if (n % i == 0) {
                curr.add(i);
                getFactorsHelper(ori, n / i, i, curr, result);
                curr.remove(curr.size() - 1);
            }
        }
    }
}

Update on 10/21/15:

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> result = new ArrayList<>();
         
        if (n <= 1) {
            return result;
        }
         
        List<Integer> curr = new ArrayList<>();
         
        getFactorsHelper(2, 1, n, curr, result);
         
        return result;
    }
     
    private void getFactorsHelper(int start, int product, int n, List<Integer> curr, List<List<Integer>> result) {
        if (start > n || product > n) {
            return;
        }
         
        if (product == n) {
            result.add(new ArrayList<Integer>(curr));
            return;
        }
         
        for (int i = start; i < n; i++) {
            if (i * product > n) {
                break;
            }
             
            if (n % (product * i) == 0) {
                curr.add(i);
                getFactorsHelper(i, product * i, n, curr, result);
                curr.remove(curr.size() - 1);
            }
        }
    }
}

Summary:
There is one trick in this problem. For each candidate factor we tried, we must make sure it is dividable by n. Thus we can avoid many unnecessary recursion calls.