LeetCode Problem 3560

Difficulty: 🟢 Easy
Topics: Math

Solution

Problem Understanding

The problem asks us to transport two logs of lengths n and m using three trucks, where each truck can carry a log of length at most k. If a log is longer than k, it must be cut into smaller pieces. Cutting a log of length x into two pieces of lengths len1 and len2 incurs a cost len1 * len2. The goal is to minimize the total cost required to make all logs fit into the trucks.

The inputs are three integers: n, m, and k. n and m are the lengths of the two logs, and k is the maximum allowed log length for each truck. The output is a single integer, representing the minimum total cutting cost.

The constraints guarantee that 1 <= n, m <= 2 * k and 2 <= k <= 105. This implies that a log may need to be cut at most once because the maximum length is only twice the truck capacity. Furthermore, the problem ensures that it is always possible to transport the logs, so we do not need to handle impossible cases.

Important edge cases include when both logs are already ≤ k, when exactly one log is > k, and when both logs exceed k but can be split optimally with minimal cost.

Approaches

The brute-force approach would consider every possible way of splitting a log that exceeds k into two valid pieces and compute the cost. This could involve iterating over all possible split points, calculating the resulting costs, and choosing the minimum combination. Although correct, this approach is unnecessary because the constraints guarantee that no log will require more than one cut.

The key insight is that for any log longer than k, there are only two ways to split it to satisfy the truck constraint: one piece of length k and the remainder. This guarantees the minimum number of cuts and minimizes the cost since any alternative would involve cutting the log into smaller pieces, increasing the product len1 * len2.

By following this observation, the optimal approach is straightforward: check each log, and if its length exceeds k, cut it once into a piece of length k and the remainder, computing the corresponding cost.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n + m)	O(1)	Iterate over all possible splits, compute cost, select minimum
Optimal	O(1)	O(1)	Cut logs exceeding k once to minimize cost

Algorithm Walkthrough

1. Initialize a variable total_cost to 0. This will hold the sum of cutting costs.
2. Check if the first log n exceeds k. If it does, compute the cost of cutting it into a piece of length k and the remainder (n - k). Add this cost to total_cost.
3. Repeat the same check for the second log m. If m > k, compute the cost of cutting it into a piece of length k and the remainder (m - k) and add it to total_cost.
4. Return the total_cost as the result.

Why it works: Each log only requires a cut if it exceeds the truck capacity k. Cutting once into a piece of length k and the remainder guarantees both pieces fit into trucks and minimizes the product len1 * len2 because splitting further increases the multiplication result. The constraints ensure that no log exceeds 2 * k, so one cut per log is always sufficient. We are given two log lengths $n$ and $m$, and three trucks. Each truck can carry exactly one log piece, and every transported piece must have length at most $k$. We are allowed to cut logs into smaller pieces, and each cut has an associated cost. Cutting a log of length $x$ into two pieces of lengths $len1$ and $len2$, where $len1 + len2 = x$, incurs cost $len1 \cdot len2$. The objective is to partition the two logs into at most three pieces total, each of length at most $k$, such that the total cutting cost is minimized.

The input consists of three integers $n, m, k$, with $2 \le k \le 10^5$ and $1 \le n, m \le 2k$. The constraint $n, m \le 2k$ is crucial because it implies that each log can require at most one cut to ensure all resulting pieces are of length at most $k$, provided we do not further subdivide beyond necessity. The problem guarantees that a valid transportation strategy always exists.

The output is a single integer: the minimum total cost required to ensure all resulting log pieces can be placed into the three trucks under the capacity constraint.

The key edge cases are when both logs already fit without cuts, when exactly one log exceeds $k$, and when both exceed $k$. The last case requires careful reasoning about feasibility under the three-truck constraint, since excessive splitting may create more than three pieces, which is disallowed.

Approaches

Brute-force approach

A direct approach is to enumerate all possible ways to cut each log into a sequence of pieces such that all pieces have length at most $k$, then select a valid partition of all resulting pieces into at most three total pieces across both logs. For each valid cutting configuration, we compute the total cost by summing all individual cut costs.

This approach is correct because it explicitly explores every valid decomposition of the logs and evaluates feasibility under the truck constraint. However, the number of possible binary partition trees for a log of length $x$ grows exponentially with the number of cuts, and even though $x \le 2k$, the combinatorial structure of all possible cut sequences makes this infeasible.

The key insight is that since $x \le 2k$, each log requires at most one “effective” cut to make all pieces valid, and the optimal cut cost has a closed-form expression. Furthermore, the total number of pieces is tightly constrained by the three-truck limit, which reduces the problem to a small constant number of structural cases.

Optimal approach

We observe that because $n, m \le 2k$, each log can be split into at most two pieces while still ensuring each piece is $\le k$. Any further splitting is unnecessary for feasibility and only increases cost and piece count.

Thus each log has two possibilities:

If $x \le k$, it requires no cut and contributes one piece.

If $k < x \le 2k$, it must be cut into exactly two pieces $i$ and $x-i$, both $\le k$. The cost function is

$$i(x-i).$$

This quadratic is concave in $i$, hence its minimum over a closed interval occurs at an endpoint. Therefore:

$$\min cost = k(x-k).$$

We then consider feasibility under at most three total pieces. Since we begin with two logs, the only valid global configurations are:

One piece + one piece (no cuts).

Two pieces + one piece (one log cut).

One piece + two pieces (one log cut).

Any configuration producing four or more pieces is invalid. Hence, at most one log may be cut.

Therefore, we compute the cost of cutting each log (if needed) and choose the best feasible option.

Complexity comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential in cut configurations	Exponential recursion depth	Enumerates all partition trees
Optimal	$O(1)$	$O(1)$	Closed-form cost + constant case analysis

Algorithm Walkthrough

We proceed by analyzing each log independently and then enforcing the global constraint on total pieces.

1. We inspect whether $n \le k$ and $m \le k$. If both are true, no cuts are required, and the total cost is zero. This follows immediately because each log already satisfies the truck capacity constraint.
2. If exactly one of $n$ or $m$ exceeds $k$, we must cut that log. We compute its cutting cost using the derived formula $k(x-k)$. The other log remains uncut. The resulting configuration produces exactly three pieces, which matches the number of trucks.
3. If both $n > k$ and $m > k$, each log individually requires at least one cut to ensure all pieces are valid. However, under the three-truck constraint, we cannot allow both logs to produce two pieces simultaneously. In this formulation, we select the cheaper cut, since only one log is allowed to be split while still respecting the total piece constraint.
4. We return the minimum cost among all feasible configurations.

Why it works

The correctness rests on two structural invariants. First, any valid decomposition of a log of length at most $2k$ that respects the constraint “all pieces $\le k$” requires at most one binary split. Second, because there are exactly two logs and at most three available trucks, the global system permits at most one log to be split without violating the total piece bound. The cost minimization therefore reduces to a constant-size optimization over at most two candidate cut operations.

Python Solution

class Solution:
    def minCuttingCost(self, n: int, m: int, k: int) -> int:
        total_cost = 0
        
        if n > k:
            total_cost += k * (n - k)
        
        if m > k:
            total_cost += k * (m - k)
        
        return total_cost

The Python implementation directly follows the algorithm steps. We initialize total_cost to 0, check each log against the capacity k, compute the cost if a cut is needed, and sum the costs. Finally, we return the total cost. def cost(x: int) -> int: if x <= k: return 0 return k * (x - k)

    # Determine feasibility under at most 3 pieces total
    n_cut = n > k
    m_cut = m > k

    if not n_cut and not m_cut:
        return 0

    if n_cut and not m_cut:
        return cost(n)

    if m_cut and not n_cut:
        return cost(m)

    # both need cutting; choose one to cut (only one can be effectively used)
    return min(cost(n), cost(m))

### Explanation

The function `cost(x)` encodes the optimal single-cut cost for a log that exceeds $k$. We then classify each log as requiring a cut or not. If neither requires cutting, the answer is zero. If exactly one requires cutting, we must pay its cost. If both require cutting, we select the cheaper one under the global constraint that only one cut operation can be effectively used while maintaining at most three transported pieces.

## Go Solution

```go
func minCuttingCost(n int, m int, k int) int64 {
    var totalCost int64 = 0

    if n > k {
        totalCost += int64(k * (n - k))
    }

    if m > k {
        totalCost += int64(k * (m - k))
    }

    return totalCost
}

In Go, we explicitly cast the multiplication to int64 to handle potential overflow, as k * (n - k) may exceed the default int size on some platforms. The logic otherwise mirrors the Python solution.

Worked Examples

Example 1: n = 6, m = 5, k = 5

1. Initialize total_cost = 0.
2. Check n = 6. Since 6 > 5, cut into 5 and 1. Cost = 5 * 1 = 5. total_cost = 5.
3. Check m = 5. Since 5 ≤ 5, no cut. total_cost remains 5.
4. Return 5.

Example 2: n = 4, m = 4, k = 6

1. Initialize total_cost = 0.

2. Check n = 4. Since 4 ≤ 6, no cut.

3. Check m = 4. Since 4 ≤ 6, no cut.

4. Return 0. cost := func(x int) int64 { if x <= k { return 0 } return int64(k) * int64(x-k) }

   nCut := n > k mCut := m > k

   if !nCut && !mCut { return 0 }

   if nCut && !mCut { return cost(n) }

   if mCut && !nCut { return cost(m) }

   if cost(n) < cost(m) { return cost(n) } return cost(m) }

### Go-specific notes

The main difference is the explicit use of `int64` to prevent overflow when computing $k(x-k)$, since both terms can be as large as $10^5$, making their product potentially exceed 32-bit integer limits. The control flow mirrors the Python implementation exactly.

## Worked Examples

### Example 1

Input:

$n = 6, m = 5, k = 5$

We evaluate each log:

For $n = 6$, since $6 > 5$, cost is:

$$k(n-k) = 5 \cdot (6-5) = 5.$$

For $m = 5$, since $5 \le 5$, cost is $0$.

We compare configurations:

Cut only $n$: valid, produces pieces $1,5$ and $5$, total 3 pieces.

Thus result is $5$.

### Example 2

Input:

$n = 4, m = 4, k = 6$

Both logs satisfy capacity constraints:

$$4 \le 6, \quad 4 \le 6.$$

No cuts are required, so total cost is $0$.

## Complexity Analysis

| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | O(1) | Only constant-time checks and multiplications are performed, independent of the input size. |
| Space | O(1) | Only a single integer variable `total_cost` is used. |

The complexity is constant because we only check two logs and potentially perform one multiplication for each.

## Test Cases

Provided examples

assert Solution().minCuttingCost(6, 5, 5) == 5 # Cut one log assert Solution().minCuttingCost(4, 4, 6) == 0 # No cuts needed

Edge cases

assert Solution().minCuttingCost(1, 1, 2) == 0 # Both logs already fit assert Solution().minCuttingCost(2, 2, 2) == 0 # Exact fit assert Solution().minCuttingCost(10, 5, 5) == 25 # Only first log needs cut assert Solution().minCuttingCost(8, 8, 5) == 18 + 15 # Both logs need cut assert Solution().minCuttingCost(5, 10, 5) == 25 # Only second log needs cut | Time | $O(1)$ | Each log is evaluated with constant arithmetic operations | | Space | $O(1)$ | No auxiliary data structures beyond constants |

The computation reduces to evaluating at most two arithmetic expressions and a constant number of comparisons, independent of input magnitude.

Test Cases

assert Solution().minCuttingCost(6, 5, 5) == 5  # example with one required cut
assert Solution().minCuttingCost(4, 4, 6) == 0  # no cuts needed
assert Solution().minCuttingCost(5, 5, 5) == 0  # boundary equality case
assert Solution().minCuttingCost(6, 6, 6) == 0  # both fit exactly
assert Solution().minCuttingCost(7, 4, 5) == 10 # only one log needs cut
assert Solution().minCuttingCost(4, 7, 5) == 10 # symmetric case
assert Solution().minCuttingCost(10, 5, 6) == 24 # cost from larger log

Test	Why
n = 6, m = 5, k = 5	Tests cutting a single log to fit
n = 4, m = 4, k = 6	Tests when no cutting is necessary
n = 1, m = 1, k = 2	Smallest possible logs
n = 2, m = 2, k = 2	Exact fit, no cut required
n = 10, m = 5, k = 5	Tests one log requiring a cut
n = 8, m = 8, k = 5	Both logs require a cut
n = 5, m = 10, k = 5	Tests cut on second log

Edge Cases

One important edge case is when both logs are already smaller than or equal to k. The algorithm correctly handles this by performing no cuts and returning 0, avoiding unnecessary calculations.

Another edge case is when one log is exactly k. This log fits perfectly into a truck, so no cut is performed. The implementation correctly uses the > comparison to skip the cut when the length equals k.

Finally, when a log is slightly larger than k but less than or equal to 2 * k, only one cut is needed. This case is crucial because splitting the log into smaller pieces incorrectly could lead to a higher cost. The algorithm ensures the minimal cost by cutting exactly once into k and the remainder. | (6,5,5) | validates single cut case | | (4,4,6) | validates no cut case | | (5,5,5) | boundary equality behavior | | (6,6,6) | both logs fit exactly | | (7,4,5) | asymmetric cut requirement | | (10,5,6) | larger cost scaling |

Edge Cases

One important edge case is when both logs exceed $k$. This is the most structurally delicate case because it appears to require two cuts, one per log, which would exceed the three-truck constraint. The implementation resolves this by selecting only one log to cut, ensuring that the total number of resulting pieces does not exceed the allowed limit.

Another edge case occurs when a log length is exactly equal to $k$. In this case, no cut is required, and treating it as a cut would incorrectly introduce unnecessary cost. The condition $x \le k$ explicitly prevents this.

A final edge case arises when $x = 2k$. This is the maximum possible length, and the optimal split is symmetric in cost terms, but the minimum cost is still correctly captured by the endpoint formula $k(x-k)$. The implementation handles this directly without enumeration, ensuring correctness even at the extreme input boundary.