Problem Description

You are given two integers n and k. Your task is to split the number n into exactly k positive integers such that when you multiply all these k integers together, the product equals n.

Among all possible ways to split n into k factors, you need to find the split where the difference between the largest and smallest number is minimized.

For example, if n = 12 and k = 3, you could split it as:

• [1, 1, 12] with product = 1 × 1 × 12 = 12, difference = 12 - 1 = 11
• [1, 2, 6] with product = 1 × 2 × 6 = 12, difference = 6 - 1 = 5
• [1, 3, 4] with product = 1 × 3 × 4 = 12, difference = 4 - 1 = 3
• [2, 2, 3] with product = 2 × 2 × 3 = 12, difference = 3 - 2 = 1

The optimal split would be [2, 2, 3] because it has the minimum difference of 1.

You should return any one valid split that achieves this minimum difference. The numbers can be returned in any order.

Intuition

To find k numbers whose product equals n with minimum difference between the largest and smallest, we need to make these numbers as close to each other as possible. Think about it: if we have factors that are very different in size (like 1 and n), the difference will be large. But if we can find factors closer to each other (like numbers around n^(1/k)), the difference will be smaller.

The key insight is that we need to explore all possible ways to factorize n into exactly k factors. This is essentially a recursive problem where at each step, we choose one factor of the remaining number and continue factorizing.

The approach pre-computes all divisors for each number up to a maximum value. For any number x, we store all its divisors in g[x]. This preprocessing step allows us to quickly access all possible factors during our search.

During the recursive search (dfs), we:

1. Start with the number n and need to find k factors
2. At each step, pick a divisor y of the current number x
3. Continue recursively with x // y and one less factor to find
4. Track the minimum and maximum factors seen so far to calculate the difference
5. When we've found all k factors (when i == 0), we check if this split gives us a better (smaller) difference than what we've seen before

The recursion explores all possible factorizations, keeping track of the split that produces the minimum difference between the largest and smallest factors. By trying all combinations systematically, we guarantee finding the optimal solution where the factors are as balanced as possible.

Solution Approach

The solution uses a depth-first search (DFS) with preprocessing to find the optimal factorization.

Preprocessing Phase:

mx = 10**5 + 1
g = [[] for _ in range(mx)]
for i in range(1, mx):
    for j in range(i, mx, i):
        g[j].append(i)

This creates a list g where g[j] contains all divisors of j. For each number i from 1 to mx-1, we add i as a divisor to all its multiples (i, 2i, 3i, ...). This gives us O(1) access to all divisors of any number during the search.

DFS Implementation: The main algorithm uses recursive DFS to explore all possible factorizations:

def dfs(i: int, x: int, mi: int, mx: int):

• i: Number of factors still needed (decrements from k-1 to 0)
• x: Current number to be factorized
• mi: Minimum factor found so far
• mx: Maximum factor found so far

Base Case: When i == 0, we've found all k factors. The last factor must be x itself. We calculate the difference between the maximum and minimum factors (considering x as well):

d = max(mx, x) - min(mi, x)

If this difference is better than our current best, we update the answer.

Recursive Case: For each divisor y of the current number x:

1. Add y to the current path at position i
2. Recursively solve for x // y with one less factor needed
3. Update the running minimum and maximum with y

Main Function:

ans = None
path = [0] * k
cur = inf
dfs(k - 1, n, inf, 0)

• Initialize path array to store the current factorization being explored
• cur tracks the best (minimum) difference found so far
• Start DFS with k-1 factors to find (since the last one is determined), starting number n, and initial min/max values

The algorithm systematically explores all valid factorizations of n into exactly k factors, keeping track of the one with the minimum difference between its largest and smallest factors.

Example Walkthrough

Let's walk through the solution with n = 12 and k = 3.

Step 1: Preprocessing First, we build the divisor list g for all numbers up to 12:

• g[12] = [1, 2, 3, 4, 6, 12] (all divisors of 12)
• g[6] = [1, 2, 3, 6]
• g[4] = [1, 2, 4]
• g[3] = [1, 3]
• g[2] = [1, 2]
• g[1] = [1]

Step 2: DFS Exploration We start with dfs(i=2, x=12, mi=inf, mx=0) and need to find 3 factors.

The DFS explores different factorization paths:

Path 1: Choose divisor 1 first

• path[2] = 1, call dfs(1, 12, 1, 1)
• Choose divisor 1 again: path[1] = 1, call dfs(0, 12, 1, 1)
• Base case: path[0] = 12
• Result: [1, 1, 12], difference = 12 - 1 = 11

Path 2: Choose divisor 2 first

• path[2] = 2, call dfs(1, 6, 2, 2)
• Choose divisor 1: path[1] = 1, call dfs(0, 6, 1, 2)
• Base case: path[0] = 6
• Result: [1, 2, 6], difference = 6 - 1 = 5

Path 3: Choose divisor 2 first

• path[2] = 2, call dfs(1, 6, 2, 2)
• Choose divisor 2: path[1] = 2, call dfs(0, 3, 2, 2)
• Base case: path[0] = 3
• Result: [2, 2, 3], difference = 3 - 2 = 1 ✓ (Best so far!)

Path 4: Choose divisor 3 first

• path[2] = 3, call dfs(1, 4, 3, 3)
• Choose divisor 1: path[1] = 1, call dfs(0, 4, 1, 3)
• Base case: path[0] = 4
• Result: [1, 3, 4], difference = 4 - 1 = 3

The algorithm continues exploring all valid combinations. Each time a complete factorization is found (when i == 0), it compares the difference with the current best (cur). The factorization [2, 2, 3] gives the minimum difference of 1, so it becomes our final answer.

Step 3: Return Result The algorithm returns [2, 2, 3] as the optimal factorization where the product 2 × 2 × 3 = 12 and the difference between max and min is minimized to 1.

Solution Implementation

Time and Space Complexity

Time Complexity: O(10^5 * d(10^5) + k * d(n)^k) where d(n) represents the number of divisors of n.

• The preprocessing step builds a graph g where g[j] contains all divisors of j. For each number i from 1 to mx-1, we iterate through its multiples up to mx-1. This takes O(mx * log(mx)) time, which equals O(10^5 * log(10^5)) or approximately O(10^5 * 17).

• The DFS explores all possible ways to decompose n into k factors. At each recursion level i, we iterate through all divisors of the current number x. In the worst case, we have k levels of recursion, and at each level, we branch into d(x) possibilities where d(x) is the number of divisors of x. The maximum number of divisors for numbers up to 10^5 is approximately 128. Therefore, the DFS has time complexity O(d(n)^k) in the worst case.

• The overall time complexity is O(10^5 * log(10^5) + d(n)^k).

Space Complexity: O(10^5 * d_avg + k) where d_avg is the average number of divisors.

• The graph g stores all divisors for each number from 1 to mx-1. Each number j stores all its divisors in g[j]. The total space used is the sum of divisors for all numbers up to 10^5, which is approximately O(10^5 * log(10^5)) or O(10^5 * 17) on average.

• The recursion stack depth is k, and we maintain a path array of size k.

• The overall space complexity is O(10^5 * log(10^5) + k).

Common Pitfalls

1. Inefficient Divisor Enumeration

One major pitfall is trying to find divisors on-the-fly during the search, which can be extremely slow:

Problematic approach:

def get_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)
    return divisors

This takes O(√n) time for each call, and when called repeatedly during DFS, it becomes a bottleneck.

Solution: Precompute all divisors using the sieve-like approach shown in the code, which gives O(1) access to divisors during search.

2. Incorrect Pruning Logic

A tempting optimization is to prune branches early when the current difference exceeds the best found so far:

Incorrect pruning:

def backtrack(position, current_product, min_factor, max_factor):
    # Wrong: This prunes too aggressively
    if max_factor - min_factor >= min_difference:
        return  # This might miss better solutions!

This is incorrect because the final factor (when position == 0) could be within the current range and still improve the overall difference.

Correct approach: Only compare differences at the base case when all factors are determined.

3. Memory Limit Issues with Large Numbers

The precomputation array has a fixed size of 10^5 + 1. If n exceeds this limit, the code will crash:

Problem scenario:

n = 10**6  # Exceeds MAX_VALUE
k = 2
# This will cause IndexError: list index out of range

Solution: Either increase MAX_VALUE appropriately or implement a hybrid approach:

def minDifference(self, n: int, k: int) -> List[int]:
    if n >= MAX_VALUE:
        # Fall back to on-the-fly divisor calculation for large n
        return self.minDifferenceWithoutPrecompute(n, k)
    # Use precomputed approach for smaller n
    ...

4. Stack Overflow with Deep Recursion

For large values of k, the recursion depth can exceed Python's default limit:

Problem:

n = 2**30  # Large number with many factors of 2
k = 30     # Deep recursion
# May cause: RecursionError: maximum recursion depth exceeded

Solution: Either increase the recursion limit or convert to iterative approach:

import sys
sys.setrecursionlimit(10000)  # Increase limit

# Or use iterative approach with explicit stack
def minDifferenceIterative(self, n: int, k: int) -> List[int]:
    stack = [(k-1, n, inf, 0, [])]
    while stack:
        position, product, min_f, max_f, path = stack.pop()
        # Process similar to recursive version

5. Overlooking Edge Cases

The code doesn't handle certain edge cases explicitly:

Edge cases to consider:

• k = 1: Should return [n]
• n = 1: Should return [1, 1, ..., 1] (k times)
• k > number of divisors of n: No valid solution exists

Enhanced solution with edge case handling:

def minDifference(self, n: int, k: int) -> List[int]:
    # Handle k = 1
    if k == 1:
        return [n]
  
    # Handle n = 1
    if n == 1:
        return [1] * k
  
    # Check if solution is possible (rough check)
    if k > 30 and n < 2**k:  # Impossible to have k factors
        return None  # Or raise exception
  
    # Continue with main algorithm...