Is Balanced K-Factor Decomposition easy or hard?

Balanced K-Factor Decomposition is generally considered a medium difficulty problem. The core math is simple, but designing an efficient recursive search with pruning requires familiarity with backtracking and divisor-based reasoning.

Balanced K-Factor Decomposition Python/Java solution

The typical implementation uses a recursive backtracking function that tracks the remaining product and the number of factors chosen. Python, Java, C++, Go, and TypeScript versions follow the same structure: compute divisors, enforce ordered selection, and prune branches when the remaining value cannot reach k factors.

How to solve Balanced K-Factor Decomposition in O(n)?

An O(n) solution is generally not possible because the task requires exploring combinations of factors. Even with pruning, the algorithm must consider multiple multiplicative partitions of n. Efficient solutions rely on backtracking with divisor filtering rather than linear-time processing.

What is the best approach for Balanced K-Factor Decomposition?

Backtracking with ordered factors and pruning is the most practical approach. Generate divisors of n, then recursively build k factors while ensuring the sequence is non-decreasing. This avoids duplicate permutations and cuts branches where the remaining product cannot produce enough factors.

Is Balanced K-Factor Decomposition asked at Google/Amazon/Meta?

Factor decomposition and multiplicative partition problems appear in interviews at large tech companies because they test recursion, pruning, and number theory reasoning. Variants involving factor combinations or product partitions have been reported in interviews at companies like Amazon and Google.

What data structure is used in Balanced K-Factor Decomposition?

The solution primarily uses recursion with a dynamic list or stack to store the current factor path. A precomputed list of divisors helps limit candidate values. The algorithm is driven by backtracking rather than specialized data structures.

What is the time complexity of Balanced K-Factor Decomposition?

The worst-case time complexity is exponential, typically expressed as O(d^k) where d is the number of divisors of n. Each recursive level tries valid divisors of the remaining value. Space complexity is O(k) due to the recursion stack and factor list.

Description Hints 3 Solution Video 12:56

3669. Balanced K-Factor Decomposition

Medium40.2% acceptance2 companies

Given two integers n and k, split the number n into exactly k positive integers such that the product of these integers is equal to n.

Return any one split in which the maximum difference between any two numbers is minimized. You may return the result in any order.

Example 1:

Input: n = 100, k = 2

Output: [10,10]

Explanation:

The split [10, 10] yields 10 * 10 = 100 and a max-min difference of 0, which is minimal.

Example 2:

Input: n = 44, k = 3

Output: [2,2,11]

Explanation:

Split [1, 1, 44] yields a difference of 43
Split [1, 2, 22] yields a difference of 21
Split [1, 4, 11] yields a difference of 10
Split [2, 2, 11] yields a difference of 9

Therefore, [2, 2, 11] is the optimal split with the smallest difference 9.

Constraints:

4 <= n <= 10⁵
2 <= k <= 5
k is strictly less than the total number of positive divisors of n.

Asked at 2 companies

Google Amazon

Topics

Hints open one at a time

Hint 1

First, compute all positive divisors of n and sort them into a list divs.

Hint 2

Use a recursive search dfs(start, picked, prod, path) that picks the next divisor from divs[start...], multiplies it into prod, and appends to path until you have k factors whose product is n.

Hint 3

During the search, keep track of the best path that minimizes max(path) – min(path) and return it.

12:56

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1class Solution:
2    def minDifference(self, n: int, k: int) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

100
2

Input

44
3

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