3560. Find Minimum Log Transportation Cost

Problem Description

You are given integers n, m, and k.

There are two logs of lengths n and m units, which need to be transported in three trucks where each truck can carry one log with length at most k units.

You may cut the logs into smaller pieces, where the cost of cutting a log of length x into logs of length len1 and len2 is cost = len1 * len2 such that len1 + len2 = x.

Return the minimum total cost to distribute the logs onto the trucks. If the logs don't need to be cut, the total cost is 0.

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Intuition

The goal is to fit two logs into three trucks, where each truck can carry at most k units in length. If both logs are already less than or equal to k, they can be loaded without any cutting or cost. If one of the logs is too long (greater than k), we need to cut it so all resulting pieces have length at most k. Since we have three trucks and two logs, at most one log needs to be split into two parts. The most efficient cut is to make one piece exactly k (so it fits on one truck), and the other piece is the remainder (x - k). The cost for such a cut, according to the problem, is k * (x - k). This way, we achieve the minimum possible cost while ensuring all logs or log pieces can be transported.

Solution Approach

First, check if both logs have lengths less than or equal to k. If so, no cuts are required, so the answer is 0.

If one of the logs is longer than k, it must be cut into two pieces so that both are at most k units in length. Since there are three trucks and two logs, only split one log: the longer one.

Let the longer log's length be x. The optimal way to cut it is to create one piece of length exactly k (maximizing the utilization of the truck) and the other piece as x - k. The cost for this cut, by the problem statement, is k * (x - k).

The solution follows these steps:

Let x = max(n, m).
If x <= k, return 0.
Otherwise, return k * (x - k).

No complex data structures or algorithms are required since this is solved with a simple mathematical calculation.

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Example Walkthrough

Suppose n = 6, m = 3, and k = 4.

Step 1: Check the lengths of the logs:

Log 1: 6 units
Log 2: 3 units Both must be transported in trucks of maximum length 4.

Step 2: Is either log longer than 4?

Log 2 (3) ≤ 4: fits in a truck
Log 1 (6) > 4: too long, needs cutting

Step 3: We cut the longer log (6 units) into two pieces:

One piece of length exactly 4 (for one truck)
The other, the remainder: 6 - 4 = 2 (fits in another truck)

Step 4: Calculate the cost:

Cost = 4 * 2 = 8

Summary: The minimum total cost is 8. The three trucks will carry logs of lengths 4, 2, and 3 units respectively.

Solution Implementation

1class Solution:
2    def minCuttingCost(self, n: int, m: int, k: int) -> int:
3        # Determine the larger dimension
4        max_side = max(n, m)
5        # If the largest side is less than or equal to k, no cutting is needed
6        if max_side <= k:
7            return 0
8        # Otherwise, the cost is k * (max_side - k)
9        return k * (max_side - k)
10

1class Solution {
2    // Finds the minimum cutting cost to reduce the longer side of a rectangle to at most k
3    public long minCuttingCost(int n, int m, int k) {
4        // Determine the longer side of the rectangle
5        int maxSide = Math.max(n, m);
6
7        // If the longer side is already less than or equal to k, no cost is required
8        if (maxSide <= k) {
9            return 0;
10        }
11        // Otherwise, cost is calculated as k times the number of cuts needed (maxSide - k)
12        return 1L * k * (maxSide - k);
13    }
14}
15

1class Solution {
2public:
3    // This function calculates the minimum cutting cost.
4    // n: length of the first side,
5    // m: length of the second side,
6    // k: maximum permitted side after cutting.
7    long long minCuttingCost(int n, int m, int k) {
8        // Find the larger of n and m
9        int maxSide = std::max(n, m);
10
11        // If the maximum side is less than or equal to k, no cut is needed
12        if (maxSide <= k) {
13            return 0;
14        }
15
16        // If a cut is needed, the cost is k * (maxSide - k)
17        // Cast to long long to avoid integer overflow
18        return 1LL * k * (maxSide - k);
19    }
20};
21

1/**
2 * Calculates the minimum cutting cost for a rectangle of size n x m,
3 * given that a cut can divide the longer side by at most k units.
4 *
5 * @param n - The height of the rectangle
6 * @param m - The width of the rectangle
7 * @param k - The maximum cut length allowed for the longer side
8 * @returns The minimum cost to cut the rectangle so that neither side exceeds k
9 */
10function minCuttingCost(n: number, m: number, k: number): number {
11    // Determine the longer side of the rectangle
12    const maxSide: number = Math.max(n, m);
13
14    // If the longer side is already within the allowed size, no cost is needed
15    if (maxSide <= k) {
16        return 0;
17    }
18
19    // Otherwise, compute the minimum cost required to cut the rectangle
20    // The cost is k for each cut, and the number of cuts is (maxSide - k)
21    return k * (maxSide - k);
22}
23
24// Example usage:
25// const result = minCuttingCost(8, 5, 4);
26// console.log(result); // Output: 16
27

Time and Space Complexity

The time complexity of the code is O(1) because it only uses a constant number of operations regardless of the input values.

The space complexity is also O(1) since it only uses a constant amount of additional space for variable storage.

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