How do you solve Balanced K-Factor Decomposition on LeetCode?

Balanced K-Factor Decomposition (LeetCode #3669) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Distributing factors evenly minimizes differences.

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

The optimal solution for Balanced K-Factor Decomposition runs in O(n) time and uses O(n) space. Finding divisors is linear, and the recursive search is efficient due to pruning.

What is the brute force solution for Balanced K-Factor Decomposition?

The brute force approach for Balanced K-Factor Decomposition is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all combinations of k positive integers that multiply to n. Check each combination to find the one with the smallest max-min difference. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Balanced K-Factor Decomposition?

Balanced K-Factor Decomposition uses the following concepts: Math, Backtracking, Number Theory. The recommended patterns to study are: Backtracking, Dynamic Programming.

What are the interview tips for Balanced K-Factor Decomposition?

Clarify constraints and requirements. Discuss edge cases and examples. Explain your thought process clearly.

What are common mistakes when solving Balanced K-Factor Decomposition?

Not considering all divisors. Ignoring the need for positive integers.

#3669

Balanced K-Factor Decomposition

Medium

MathBacktrackingNumber TheoryBacktrackingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Distribute the factors of n as evenly as possible among k integers to minimize the maximum difference.

⚙️

Algorithm

3 steps

1Step 1: Find all divisors of n and sort them.
2Step 2: Use a recursive function to select k divisors such that their product equals n.
3Step 3: Ensure the selected divisors are as close in value as possible.

solution.py14 lines

1def balanced_k_factor(n, k):
2    def dfs(start, picked, prod, path):
3        if len(path) == k:
4            if prod == n:
5                return path
6            return None
7        for i in range(start, len(divs)):
8            if prod * divs[i] > n:
9                break
10            res = dfs(i, picked + 1, prod * divs[i], path + [divs[i]])
11            if res:
12                return res
13    divs = sorted([i for i in range(1, n + 1) if n % i == 0])
14    return dfs(0, 0, 1, [])

ℹ

Complexity note: Finding divisors is linear, and the recursive search is efficient due to pruning.

1Distributing factors evenly minimizes differences.
2Recursive backtracking can efficiently explore combinations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.