Solution Explanation for Factor Combinations

The problem asks to find all possible combinations of factors for a given integer n. The factors must be in the range [2, n-1]. This is a classic backtracking problem.

Approach:

The solution uses a depth-first search (DFS) approach implemented through a recursive helper function (dfs). The core idea is to explore all possible factor combinations systematically.

1. Base Case: If the current number (n) is 1, it means we've found a valid combination of factors. We append the current combination (tracked in t) along with the remaining n to ans.

2. Recursive Step: The algorithm iterates through potential factors (j) from i up to the square root of n. i starts at 2 and is passed recursively. This optimization avoids duplicate factor pairs (e.g., for 12, we consider 2 and 6, but not 6 and 2).

3. Factor Found: If j is a factor of n (n % j == 0), we add j to the current combination (t), recursively call dfs with n/j (the remaining part after dividing by the factor) and j (to prevent considering factors smaller than the previously found ones), and then remove j from t (backtracking).

Time Complexity: O(N^(N/2)), where N is the input number. In the worst-case scenario (a number with many prime factors), the algorithm explores a large number of combinations. The complexity is exponential because the number of factor combinations can grow very rapidly.

Space Complexity: O(log N + K), where log N represents space used for recursive calls (depth of recursion is at most log₂(n)), and K is the maximum length of a factor combination.

Code Explanation (Python):

class Solution:
    def getFactors(self, n: int) -> List[List[int]]:
        def dfs(n, i):
            if t:  # If t is not empty, we have a valid combination
                ans.append(t + [n]) # append the combination to the answer
 
            j = i #start from i to prevent smaller factors
            while j * j <= n: # optimization: only check up to sqrt(n)
                if n % j == 0:  # If j is a factor
                    t.append(j)  # Add j to the current combination
                    dfs(n // j, j)  # Recursive call with the remaining part
                    t.pop()  # Backtrack: remove j for exploring other combinations
                j += 1  # Move to the next potential factor
 
        t = []  # List to store current combination
        ans = []  # List to store all combinations
        dfs(n, 2)  # Start DFS from 2
        return ans
 

The other languages (Java, C++, Go) follow the same logic, employing the same backtracking algorithm. The only difference lies in the syntax specific to each language. The core data structures (List, vector, etc.) and recursive function structure are analogous across all implementations.