Problem Description

You have two logs with lengths n and m units that need to be transported using exactly three trucks. Each truck can carry one log with a maximum length of k units.

You can cut the logs into smaller pieces to fit them onto the trucks. When you cut a log of length x into two pieces with lengths len1 and len2 (where len1 + len2 = x), the cutting cost is len1 * len2.

Your goal is to find the minimum total cost to distribute all the log pieces onto the three trucks such that:

• Each truck carries at most one log piece
• Each log piece has length at most k units
• All original log material is transported (the sum of all pieces equals n + m)

If the logs can be loaded without any cutting, the total cost is 0.

For example, if you have logs of lengths n = 10 and m = 5, and trucks with capacity k = 8:

• The first log (length 10) exceeds the truck capacity, so it must be cut
• One way is to cut it into pieces of length 8 and 2, with cost 8 * 2 = 16
• The second log (length 5) fits without cutting
• The three pieces (8, 2, 5) can each go on one truck
• The total cost would be 16

The solution recognizes that with only 3 trucks available for 2 logs, at most one log needs to be cut (into exactly 2 pieces). If both logs fit within the capacity k, no cutting is needed. Otherwise, the longer log must be cut, and the optimal cut is to make one piece exactly k units long, resulting in a cost of k * (x - k) where x is the length of the longer log.

Intuition

Let's think about what we're working with: 2 logs and 3 trucks. This means we can transport at most 3 pieces total.

If both logs already fit within the truck capacity k, we don't need to cut anything - we just load the 2 logs onto 2 trucks and leave one truck empty. Cost = 0.

Now, what if at least one log is longer than k? Since we have 3 trucks for 2 logs, we can cut at most one log into exactly 2 pieces (giving us 3 pieces total to fill our 3 trucks). We can't cut both logs because that would give us at least 4 pieces, but we only have 3 trucks.

So the question becomes: if we need to cut one log, which one should we cut and how?

Obviously, we should cut the longer log (let's call its length x where x > k). But where should we make the cut?

For a log of length x that needs to be cut into two pieces that both fit within capacity k, we need both pieces to be at most k units long. If we cut it into pieces of length a and b where a + b = x, the cost is a * b.

To minimize a * b subject to:

• a + b = x (pieces sum to original length)
• a ≤ k and b ≤ k (both pieces must fit in trucks)

Since x > k, we know that at least one piece must be less than k. The key insight is that to minimize the product a * b when the sum is fixed, we want the values to be as far apart as possible (unlike maximizing a product, where we'd want them equal).

The most extreme split that still satisfies our constraints is to make one piece exactly k units and the other x - k units. This gives us a cost of k * (x - k).

This is why the solution simply checks if the longer log exceeds k, and if so, returns k * (x - k) as the minimum cost.

Learn more about Math patterns.

Solution Approach

The implementation is straightforward once we understand the mathematical insight:

1. Identify the longer log: First, we find the maximum length between the two logs using x = max(n, m). This is the log that potentially needs cutting.

2. Check if cutting is needed: We compare the longer log's length x with the truck capacity k:

   • If x <= k, both logs fit within the truck capacity. We can load them directly without any cuts, so we return 0.
   • If x > k, the longer log exceeds the truck capacity and must be cut.

3. Calculate the minimum cutting cost: When cutting is required, we cut the longer log into two pieces:

   • One piece of length k (the maximum that fits in a truck)
   • Another piece of length x - k (the remainder)
   • The cutting cost is k * (x - k)

The complete solution in one line:

return 0 if x <= k else k * (x - k)

This works because:

• With 3 trucks and 2 logs, we have enough capacity to handle one log being split into 2 pieces
• The shorter log (with length min(n, m)) will always fit in a truck since if it didn't, we'd need to cut both logs into at least 4 pieces total, which exceeds our 3 truck capacity
• The optimal cut for minimizing cost when splitting a log of length x into pieces that fit within capacity k is always to make one piece exactly k units long

The time complexity is O(1) as we only perform constant-time operations. The space complexity is also O(1) as we only use a single variable to store the maximum length.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: n = 7, m = 11, k = 6

Step 1: Identify the longer log

• We have two logs with lengths 7 and 11
• The longer log has length x = max(7, 11) = 11

Step 2: Check if cutting is needed

• Compare x with truck capacity k: Is 11 ≤ 6?
• No! Since 11 > 6, the longer log exceeds truck capacity and must be cut

Step 3: Calculate the minimum cutting cost

• We need to cut the log of length 11 into two pieces
• To minimize cost, we make one piece exactly k = 6 units long
• The other piece will be 11 - 6 = 5 units long
• Both pieces now fit in trucks (6 ≤ 6 ✓ and 5 ≤ 6 ✓)
• The cutting cost is 6 × 5 = 30

Final arrangement:

• Truck 1: carries a piece of length 6 (from the cut log)
• Truck 2: carries a piece of length 5 (from the cut log)
• Truck 3: carries the entire second log of length 7... wait, that won't work!

Actually, let me reconsider. The second log has length 7, which also exceeds k = 6. But we only have 3 trucks total. If we cut both logs, we'd have at least 4 pieces, but only 3 trucks to carry them. This scenario is impossible to solve with the given constraints.

Let's use a valid example instead: n = 7, m = 4, k = 6

Step 1: Identify the longer log

• The longer log has length x = max(7, 4) = 7

Step 2: Check if cutting is needed

• Compare x with k: Is 7 ≤ 6?
• No! Since 7 > 6, we need to cut

Step 3: Calculate the minimum cutting cost

• Cut the log of length 7 into pieces of length 6 and 1
• Cutting cost = 6 × 1 = 6

Final arrangement:

• Truck 1: piece of length 6 (from cutting the 7-unit log)
• Truck 2: piece of length 1 (from cutting the 7-unit log)
• Truck 3: the entire 4-unit log (fits without cutting)
• Total cost: 6

This demonstrates why the formula k × (x - k) gives us the minimum cost when cutting is necessary.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a fixed number of operations regardless of the input size. It only executes:

• One max() operation between two numbers
• One comparison (x <= k)
• At most one arithmetic operation (k * (x - k))

All these operations take constant time.

The space complexity is O(1) because the function only uses a constant amount of extra space. It stores:

• One variable x to hold the maximum value
• The function parameters n, m, and k which don't count as extra space
• No additional data structures that grow with input size

The function returns either 0 or the result of a simple arithmetic calculation without using any additional memory that scales with the input.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Cut Both Logs

A common mistake is thinking that we might need to cut both logs to minimize cost. This leads to unnecessarily complex solutions that try to optimize cuts across both logs.

Why it's wrong: With only 3 trucks available and 2 logs to transport, we can have at most 3 pieces total. This means at most one log can be cut (into 2 pieces), while the other must remain whole.

Incorrect approach:

def minCuttingCost(self, n: int, m: int, k: int) -> int:
    # Wrong: Trying to optimize cuts for both logs
    if n <= k and m <= k:
        return 0
  
    cost = 0
    if n > k:
        cost += k * (n - k)
    if m > k:
        cost += k * (m - k)
  
    return cost  # This assumes we can cut both logs, which we can't!

2. Misunderstanding the Cutting Cost Formula

Some might incorrectly calculate the cutting cost as the sum of pieces or use the wrong multiplication.

Wrong formulas:

• Using x + k instead of k * (x - k)
• Using (x - k) * (x - k) thinking both pieces have the same length
• Using x * k directly

Incorrect implementation:

def minCuttingCost(self, n: int, m: int, k: int) -> int:
    x = max(n, m)
    if x <= k:
        return 0
    return x - k  # Wrong: This is not the cutting cost!

3. Over-optimizing the Cut Position

Some solutions try to find the "optimal" cut position by minimizing the product len1 * len2 subject to both pieces being ≤ k.

Why it's unnecessary: When we must cut a log of length x > k, the constraint that each piece must be ≤ k forces us to cut it as k and (x - k). There's no other valid way to cut it with both pieces fitting in trucks.

Overly complex approach:

def minCuttingCost(self, n: int, m: int, k: int) -> int:
    x = max(n, m)
    if x <= k:
        return 0
  
    # Unnecessary optimization attempt
    min_cost = float('inf')
    for cut_pos in range(1, x):
        if cut_pos <= k and (x - cut_pos) <= k:
            min_cost = min(min_cost, cut_pos * (x - cut_pos))
  
    return min_cost  # This will always result in k * (x - k) anyway

4. Forgetting Edge Cases

Not handling the case where no cutting is needed (when both logs fit within capacity).

Incomplete solution:

def minCuttingCost(self, n: int, m: int, k: int) -> int:
    x = max(n, m)
    return k * (x - k)  # Wrong: Returns negative cost when x <= k!

Correct Solution: The key insight is recognizing the problem's constraints lead to a simple, deterministic solution:

def minCuttingCost(self, n: int, m: int, k: int) -> int:
    x = max(n, m)
    return 0 if x <= k else k * (x - k)