How do you solve Find Minimum Log Transportation Cost on LeetCode?

Find Minimum Log Transportation Cost (LeetCode #3560) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Logs fitting without cuts saves cost.

What is the time complexity of Find Minimum Log Transportation Cost?

The optimal solution for Find Minimum Log Transportation Cost runs in O(1) time and uses O(1) space. Constant time complexity since we are only performing a few comparisons and calculations regardless of input size.

What is the brute force solution for Find Minimum Log Transportation Cost?

The brute force approach for Find Minimum Log Transportation Cost is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible ways to cut the logs and calculate the costs. This method is exhaustive and guarantees finding the minimum cost but is inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Find Minimum Log Transportation Cost?

Find Minimum Log Transportation Cost uses the following concepts: Math. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Find Minimum Log Transportation Cost?

Clarify constraints before coding. Discuss edge cases with the interviewer. Explain your thought process clearly.

What are common mistakes when solving Find Minimum Log Transportation Cost?

Overcomplicating the cut calculations. Ignoring cases where no cuts are needed.

#3560

Find Minimum Log Transportation Cost

Easy

MathGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

Directly assess if logs need cutting based on their lengths compared to k. If both logs exceed k, calculate the minimal cut needed to fit them into trucks.

⚙️

Algorithm

3 steps

1Step 1: If both n and m are less than or equal to k, return 0.
2Step 2: If both n and m are greater than k, calculate the minimal cut cost as (n - k) * (m - k).
3Step 3: If only one log exceeds k, cut it to fit and return the cost.

solution.py6 lines

1def min_cost_optimal(n, m, k):
2    if n <= k and m <= k:
3        return 0
4    if n > k and m > k:
5        return (n - k) * (m - k)
6    return 0

ℹ

Complexity note: Constant time complexity since we are only performing a few comparisons and calculations regardless of input size.

1Logs fitting without cuts saves cost.
2Minimize cuts by assessing lengths directly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.