[leetcode] Factor Combinations


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:

  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]
]

DFS approach.

When writing DFS algorithm, be aware of symmetric, which is, we must ‘retrieve spot’ after we backtrack from child nodes.

For example,

thisRes.push_back(i);
DFS(n / i, res, thisRes);
thisRes.pop_back(); // pop i

and

if(thisRes.size() > 0){ // single factor is not allowed due to problem specification.
    thisRes.push_back(n);
    res.push_back(thisRes);
    thisRes.pop_back();
}

 

 

class Solution {
public:
    vector<vector<int>> getFactors(int n) {
        vector<vector<int> > res;
        vector<int> thisRes;
        DFS(n, res, thisRes);
        return res;
    }
    
    void DFS(int n, vector<vector<int> >& res, vector<int>& thisRes){
        int i;
        i = thisRes.size() == 0? 2: thisRes.back();// i is greater than last element in thisRes. This is to avoid duplicate and make thisRes in ascending order.
        for(; i <= sqrt(n); i++){
            if(n % i == 0){
                //i is a factor of n
                thisRes.push_back(i);
                DFS(n / i, res, thisRes);
                thisRes.pop_back(); // pop i
            }
        } 
        if(thisRes.size() > 0){
            thisRes.push_back(n);
            res.push_back(thisRes);
            thisRes.pop_back();
        }
    }
};

 

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