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254. Factor Combinations πŸ”’

Description

Numbers can be regarded as the product of their factors.

  • For example, 8 = 2 x 2 x 2 = 2 x 4.

Given an integer n, return all possible combinations of its factors. You may return the answer in any order.

Note that the factors should be in the range [2, n - 1].

 

Example 1:

Input: n = 1
Output: []

Example 2:

Input: n = 12
Output: [[2,6],[3,4],[2,2,3]]

Example 3:

Input: n = 37
Output: []

 

Constraints:

  • 1 <= n <= 107

Solutions

Solution 1

Thinking

Factorizations of \(n\) with at least two factors should be generated in nondecreasing order to avoid permutations. Enumerate factors from the current minimum \(i\) up to \(\sqrt{n}\).

\(dfs(n,i)\) records the chosen factors plus leftover \(n\), then tries each \(j\ge i\) that divides \(n\) and recurses on \(n/j\).

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class Solution:
    def getFactors(self, n: int) -> List[List[int]]:
        def dfs(n, i):
            if t:
                ans.append(t + [n])
            j = i
            while j * j <= n:
                if n % j == 0:
                    t.append(j)
                    dfs(n // j, j)
                    t.pop()
                j += 1

        t = []
        ans = []
        dfs(n, 2)
        return ans

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class Solution {
    private List<Integer> t = new ArrayList<>();
    private List<List<Integer>> ans = new ArrayList<>();

    public List<List<Integer>> getFactors(int n) {
        dfs(n, 2);
        return ans;
    }

    private void dfs(int n, int i) {
        if (!t.isEmpty()) {
            List<Integer> cp = new ArrayList<>(t);
            cp.add(n);
            ans.add(cp);
        }
        for (int j = i; j <= n / j; ++j) {
            if (n % j == 0) {
                t.add(j);
                dfs(n / j, j);
                t.remove(t.size() - 1);
            }
        }
    }
}

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class Solution {
public:
    vector<vector<int>> getFactors(int n) {
        vector<int> t;
        vector<vector<int>> ans;
        function<void(int, int)> dfs = [&](int n, int i) {
            if (t.size()) {
                vector<int> cp = t;
                cp.emplace_back(n);
                ans.emplace_back(cp);
            }
            for (int j = i; j <= n / j; ++j) {
                if (n % j == 0) {
                    t.emplace_back(j);
                    dfs(n / j, j);
                    t.pop_back();
                }
            }
        };
        dfs(n, 2);
        return ans;
    }
};

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func getFactors(n int) [][]int {
    t := []int{}
    ans := [][]int{}
    var dfs func(n, i int)
    dfs = func(n, i int) {
        if len(t) > 0 {
            ans = append(ans, append(slices.Clone(t), n))
        }
        for j := i; j <= n/j; j++ {
            if n%j == 0 {
                t = append(t, j)
                dfs(n/j, j)
                t = t[:len(t)-1]
            }
        }
    }
    dfs(n, 2)
    return ans
}

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