Scale Engineer

Find Minimum Log Transportation Cost

Easy
#3170Time: O(n + m) - The algorithm involves two loops that iterate up to `n-1` and `m-1` times in the worst case.Space: O(1) - The algorithm uses a constant amount of extra space.
Patterns
Math

Prompt

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.

 

Example 1:

Input: n = 6, m = 5, k = 5

Output: 5

Explanation:

Cut the log with length 6 into logs with length 1 and 5, at a cost equal to 1 * 5 == 5. Now the three logs of length 1, 5, and 5 can fit in one truck each.

Example 2:

Input: n = 4, m = 4, k = 6

Output: 0

Explanation:

The two logs can fit in the trucks already, hence we don't need to cut the logs.

 

Constraints:

  • 2 <= k <= 105
  • 1 <= n, m <= 2 * k
  • The input is generated such that it is always possible to transport the logs.

Approaches

2 approaches with complexity analysis and trade-offs.

This approach attempts to find the minimum cost by exploring the simplest cutting scenarios: making at most one cut in total. This results in either two or three final log pieces. If the original logs n and m are both no longer than k, no cuts are needed, and the cost is zero. Otherwise, the algorithm exhaustively tries every possible single cut on log n (if m <= k) and log m (if n <= k), checking if the resulting three pieces can all fit into trucks. It keeps track of the minimum cost found.

Algorithm

  1. Initialize a variable min_cost to a very large value.
  2. First, consider the case with no cuts. If n <= k and m <= k, the logs can be transported without any cuts. The cost is 0. This is the optimal solution in this case.
  3. If the no-cut case is not possible, we must make at least one cut. This approach assumes we only need to make one total cut.
  4. Scenario A: Cut log n. Iterate through all possible integer cut points i from 1 to n-1.
    • The pieces are i, n-i, and the uncut log m.
    • Check if all three pieces are valid, i.e., i <= k, n-i <= k, and m <= k.
    • If they are valid, calculate the cost i * (n-i) and update min_cost = min(min_cost, i * (n-i)).
  5. Scenario B: Cut log m. Iterate through all possible integer cut points j from 1 to m-1.
    • The pieces are n, j, and m-j.
    • Check if all three pieces are valid, i.e., n <= k, j <= k, and m-j <= k.
    • If they are valid, calculate the cost j * (m-j) and update min_cost = min(min_cost, j * (m-j)).
  6. Return the final min_cost.

Walkthrough

The core idea is to check all possibilities that result in three or fewer pieces. This happens in two main situations:

  1. Zero cuts: If n <= k and m <= k, we have two pieces that fit in two of the three trucks. The cost is 0.
  2. One cut: If one log is larger than k but the other is not (e.g., n > k and m <= k), we can cut the larger log. We iterate through all possible cut points for the larger log and find the minimum cost among valid cuts. A cut of log n at i is valid if the resulting pieces i and n-i, along with log m, are all less than or equal to k.

This method is essentially a brute-force search over all single-cut possibilities.

class Solution {    public long minCost(int n, int m, int k) {        if (n <= k && m <= k) {            return 0;        }         long minCost = Long.MAX_VALUE;         // Case 1: Cut log n, m is not cut        if (m <= k) {            for (int i = 1; i < n; i++) {                if (i <= k && (n - i) <= k) {                    minCost = Math.min(minCost, (long)i * (n - i));                }            }        }         // Case 2: Cut log m, n is not cut        if (n <= k) {            for (int i = 1; i < m; i++) {                if (i <= k && (m - i) <= k) {                    minCost = Math.min(minCost, (long)i * (m - i));                }            }        }                // This approach doesn't handle n > k and m > k,         // but the optimal solution will cover it.        // If minCost is still MAX_VALUE, it means n>k and m>k, which this approach can't solve.        // However, the problem guarantees a solution exists.        // The optimal cut for the single-cut case is always (L-k, k), so we can optimize the loops.        if (n > k && m <= k) {            minCost = Math.min(minCost, (long)k * (n - k));        }        if (m > k && n <= k) {            minCost = Math.min(minCost, (long)k * (m - k));        }         return minCost;    }}

Complexity

Time

O(n + m) - The algorithm involves two loops that iterate up to `n-1` and `m-1` times in the worst case.

Space

O(1) - The algorithm uses a constant amount of extra space.

Trade-offs

Pros

  • Simple to understand and implement.

  • Correctly solves cases where at most one log is larger than k.

Cons

  • This approach is based on a flawed model that only considers a maximum of three final pieces. It fails to find a solution for the case where both n > k and m > k, as this would require at least two cuts, resulting in four pieces, which contradicts the model's assumption.

  • The time complexity is dependent on the lengths of the logs, which can be large.

Solutions

class Solution {public  long minCuttingCost(int n, int m, int k) {    int x = Math.max(n, m);    return x <= k ? 0 : 1L * k * (x - k);  }}

Video walkthrough

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