Skip to main content
Talentd

Balanced K-Factor Decomposition - Solution & Explanation

MediumMathBacktrackingNumber Theory6 min readAsked at: Amazon, Google
Practice this problem

Problem Statement

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.

Approach Overview

Problem Overview: You are given an integer n and a target count k. The task is to decompose n into exactly k integer factors whose product equals n while keeping the factors as balanced as possible (no unnecessary skew toward very small or very large factors).

Approach 1: Exhaustive Factor Combination (Brute Force) (Time: O(d^k), Space: O(k))

Start by generating all divisors of n. Then try every possible combination of k divisors whose product equals n. A recursive search multiplies chosen factors and tracks how many have been used so far. If the product exceeds n or the count exceeds k, the branch stops early. This method is straightforward but explores many redundant combinations, especially when n has many divisors.

Approach 2: Backtracking with Ordered Factors and Pruning (Time: O(d^k) worst case, typically much smaller; Space: O(k))

A better strategy uses backtracking combined with properties from number theory. Instead of trying all divisor permutations, enforce a non‑decreasing order of factors. At each recursion step, iterate through divisors starting from the previous factor to avoid duplicate arrangements. Divide the remaining product by the chosen factor and continue searching. If the remaining value cannot produce enough factors to reach k, prune the branch immediately.

The key insight is that factor combinations behave like multiplicative partitions. Ordering the factors and shrinking the remaining product drastically reduces the search space. This approach also keeps factors naturally balanced because extremely small factors quickly force large remaining values, which fail the pruning checks.

Precomputing divisors of n in O(√n) time further speeds up the recursion. Each step only explores valid divisors of the remaining number, making the search practical even when the theoretical worst case is exponential.

Recommended for interviews: Interviewers expect the optimized backtracking approach. The brute force solution shows that you understand factor generation, but the ordered recursion with pruning demonstrates deeper understanding of math structure and efficient search space reduction.

Solution

Code

PythonJavaC++GoTypeScript

Python

Java

C++

Go

TypeScript

Try this approach in the editor →

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Exhaustive Factor CombinationO(d^k)O(k)Useful for understanding multiplicative partitions or when n has very few divisors
Backtracking with Ordered Factors and PruningO(d^k) worst caseO(k)General solution. Avoids duplicate permutations and prunes impossible branches early

Video Solution

Balanced K-Factor Decomposition | LeetCode 3669 | Weekly Contest 465 • Sanyam IIT Guwahati • 1,573 views views

Watch 9 more video solutions →

Frequently Asked Questions

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.

Ready to solve this problem?

Practice Balanced K-Factor Decomposition with our built-in code editor and test cases.

Practice on FleetCode
Previous

Add Two Numbers

Next

Reverse Integer