Is Factor Combinations easy or hard?

Factor Combinations is generally classified as a Medium problem. The challenge comes from designing a recursive backtracking process that avoids duplicates and prunes the search using sqrt(n). Developers familiar with DFS-style enumeration usually solve it quickly.

Factor Combinations Python/Java solution

Python, Java, C++, and Go implementations follow the same pattern: maintain a list for the current path, iterate from a starting factor to sqrt(n), check divisibility, append the factor, recurse on n / factor, then backtrack. Each valid divisor also forms a combination with its complementary factor.

How to solve Factor Combinations in O(n)?

An exact O(n) solution generally doesn't exist because the algorithm must generate every valid factor combination. The practical optimization limits divisor checks to sqrt(n) and uses backtracking to explore only valid branches. This reduces unnecessary exploration while still enumerating all results.

What is the best approach for Factor Combinations?

Backtracking with a start index and a sqrt(n) bound is the standard solution. It recursively builds factor combinations while ensuring factors appear in non‑decreasing order. This avoids duplicates and reduces the search space by only checking divisors up to sqrt(n).

Is Factor Combinations asked at Google/Amazon/Meta?

Factor combination generation appears in interviews at companies like Google, Amazon, and Meta as a recursion or backtracking exercise. It tests divisor logic, pruning strategies, and the ability to avoid duplicate permutations during recursive enumeration.

What data structure is used in Factor Combinations?

The main structure is a dynamic list or array used as a recursion path to store the current factor combination. The algorithm also relies on recursion (call stack) to explore deeper factor partitions. No advanced structures like heaps or hash maps are required.

What is the time complexity of Factor Combinations?

The typical backtracking solution runs in roughly O(sqrt(n) * k), where k is the number of valid factor combinations produced. Each recursion step checks divisibility only up to sqrt of the remaining number. Space complexity is O(k) due to recursion depth and the temporary path used to build combinations.

254. Factor Combinations - Practice Online

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