Find Minimum Log Transportation Cost - Practice Coding Problems

Find Minimum Log Transportation Cost - Problem

Math Easy

Log Transportation Challenge

You're a logistics manager for a lumber company with two logs of lengths n and m units that need to be transported using three trucks. Each truck can carry at most k units of wood.

Here's the challenge: if a log is too long for a truck, you must cut it into smaller pieces. However, cutting comes at a cost! When you cut a log of length x into two pieces of lengths len1 and len2, the cutting cost is len1 × len2 (where len1 + len2 = x).

Goal: Find the minimum total cost to cut and distribute both logs across the three trucks. If no cutting is needed, the cost is 0.

Example: If you have logs of length 8 and 6, with truck capacity 5, you might cut the first log into pieces of 3 and 5 (cost = 3×5 = 15), then load them as: [3], [5], [6].

Input & Output

example_1.py — Python

$ Input: n = 8, m = 6, k = 5

› Output: 24

💡 Note: Log n=8 needs to be cut (8 > 5). Optimal cut at position 3: pieces [3,5], cost = 3×5 = 15. Log m=6 needs to be cut (6 > 5). Optimal cut at position 3: pieces [3,3], cost = 3×3 = 9. Total pieces: [3,5,3,3] can fit in 3 trucks. Total cost = 15 + 9 = 24.

example_2.py — Python

$ Input: n = 4, m = 3, k = 5

› Output: 0

💡 Note: Both logs fit in trucks without cutting (4 ≤ 5 and 3 ≤ 5). Load as [4], [3], [empty]. No cutting needed, so cost = 0.

example_3.py — Python

$ Input: n = 10, m = 8, k = 3

› Output: 42

💡 Note: Log n=10 needs 4 pieces (⌈10/3⌉=4): cut into [3,3,3,1], cost = 3×7 + 3×4 + 3×1 = 21+12+3 = 36. Log m=8 needs 3 pieces (⌈8/3⌉=3): cut into [3,3,2], cost = 3×5 + 3×2 = 15+6 = 21. But we need 7 trucks total, we only have 3. This seems impossible with given constraints.

Constraints

1 ≤ n, m ≤ 10⁴
1 ≤ k ≤ 10⁴
You have exactly 3 trucks available
The sum of pieces from both logs must fit in 3 trucks

Visualization

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Understanding the Visualization

Assess the Logs

Check if logs n and m fit in truck capacity k

Plan the Cuts

For logs that don't fit, determine minimum pieces needed

Optimize Cut Positions

Choose cut positions that minimize cost (balanced pieces)

Load the Trucks

Verify all pieces fit in 3 trucks and calculate total cost

Key Takeaway

🎯 Key Insight: Use mathematical optimization instead of brute force - analyze minimum pieces needed and calculate optimal cut positions using formulas rather than trying all combinations.

Asked in

A Amazon 45 G Google 38 M Microsoft 32 F Meta 28

The optimal approach uses mathematical optimization to determine the minimum cutting cost. For each log that exceeds truck capacity, calculate the minimum pieces needed and find optimal cut positions that minimize the product cost. The key insight is that balanced cuts minimize cost - when cutting into 2 pieces, cut near the middle; when cutting into 3 pieces, make pieces as equal as possible. Time complexity is O(1) as it uses direct mathematical formulas.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Cuts)	O(2^n × 2^m)	O(2^n × 2^m)	Try every possible way to cut both logs and find minimum cost
Mathematical Optimization	O(1)	O(1)	Use mathematical analysis to find optimal cutting points

Brute Force (Try All Cuts) — Algorithm Steps

Generate all possible ways to cut log n into pieces ≤ k
Generate all possible ways to cut log m into pieces ≤ k
Try all combinations of cuts for both logs
Check if pieces can fit in 3 trucks
Calculate cost for each valid combination
Return minimum cost found

Visualization

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Step-by-Step Walkthrough

Generate All Cuts for Log 1

Try cutting at every position

Generate All Cuts for Log 2

Try cutting at every position

Test Combinations

Check if pieces fit in 3 trucks

Find Minimum

Select combination with lowest cost

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int min(int a, int b) {
    return a < b ? a : b;
}

int minimumCostRecursive(int length, int k, int* memo, int maxLen) {
    if (length <= k) return 0;
    if (memo[length] != -1) return memo[length];
    
    int minCost = INT_MAX;
    for (int i = 1; i < length; i++) {
        if (i <= k && (length - i) <= k) {
            minCost = min(minCost, i * (length - i));
        } else {
            int leftCost = minimumCostRecursive(i, k, memo, maxLen);
            int rightCost = minimumCostRecursive(length - i, k, memo, maxLen);
            if (leftCost != INT_MAX && rightCost != INT_MAX) {
                minCost = min(minCost, leftCost + rightCost + i * (length - i));
            }
        }
    }
    
    memo[length] = minCost;
    return minCost;
}

int minimumCost(int n, int m, int k) {
    int maxLen = n > m ? n : m;
    int* memo = (int*)malloc((maxLen + 1) * sizeof(int));
    
    for (int i = 0; i <= maxLen; i++) {
        memo[i] = -1;
    }
    
    int costN = minimumCostRecursive(n, k, memo, maxLen);
    int costM = minimumCostRecursive(m, k, memo, maxLen);
    
    free(memo);
    
    if (costN == INT_MAX || costM == INT_MAX) return 0;
    return costN + costM;
}

int main() {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    printf("%d\n", minimumCost(n, m, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n × 2^m)

We generate all possible cuts for both logs, leading to exponential combinations

⚠ Quadratic Growth

Space Complexity

O(2^n × 2^m)

Need to store all possible cutting combinations

⚠ Quadratic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Cuts) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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