Find Minimum Log Transportation Cost

Find Minimum Log Transportation Cost - Problem

Math Easy

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.

Input & Output

Example 1 — Basic Case

$ Input: n = 7, m = 4, k = 3

› Output: 18

💡 Note: Log n=7: Cut into pieces [3,4]. Cost = 3×4 = 12. Then cut 4 into [3,1]. Cost = 3×1 = 3. Total for log 7: 15. Log m=4: Cut into pieces [3,1]. Cost = 3×1 = 3. Total = 15 + 3 = 18.

Example 2 — No Cuts Needed

$ Input: n = 3, m = 2, k = 5

› Output: 0

💡 Note: Both logs fit in trucks without cutting since 3 ≤ 5 and 2 ≤ 5. Total cost = 0.

Example 3 — One Log Needs Cutting

$ Input: n = 8, m = 3, k = 4

› Output: 16

💡 Note: Log n=8: Cut into [4,4]. Cost = 4×4 = 16. Log m=3: No cut needed. Total = 16.

Constraints

1 ≤ n, m, k ≤ 1000
We have exactly 3 trucks available
Each truck can carry at most one log

Visualization

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Asked in

G Google 25 M Microsoft 20 A Amazon 15

The key insight is to cut logs optimally by removing pieces of maximum size k from the end. The mathematical approach gives O(1) time per log. Best approach uses the formula: while remaining > k, cut piece of size k with cost = k × remaining_after_cut. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force - Try All Cut Combinations

⏱️ Time: O(2^(n+m)) Space: O(n+m)

For each log, try all possible ways to cut it into pieces that fit in trucks. Calculate the total cost for each combination and return the minimum.

Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Key insight: to minimize cost, cut the log into pieces as close to k as possible. If a log needs cuts, the optimal strategy is to make all pieces equal to k except possibly one remainder piece.

Memoization - Cache Results

⏱️ Time: O(n² + m²) Space: O(n + m)

Same recursive approach as brute force but cache results to avoid redundant calculations. Store minimum cost for each log length in a memo table.

Brute Force - Try All Cut Combinations — Algorithm Steps

Try all ways to cut log n into pieces ≤ k
Try all ways to cut log m into pieces ≤ k
Calculate cost for each combination
Return minimum total cost

Visualization

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Step-by-Step Walkthrough

Try All Cuts

Recursively try cutting at each position

Calculate Cost

For each cut: cost = left * right + sub-costs

Find Minimum

Return minimum cost among all possibilities

Code -

solution.c — C

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

int minCostToCut(int length, int k) {
    if (length <= k) {
        return 0;
    }
    
    int minCost = INT_MAX;
    for (int cutPos = 1; cutPos < length; cutPos++) {
        int left = cutPos;
        int right = length - cutPos;
        int cost = left * right + minCostToCut(left, k) + minCostToCut(right, k);
        if (cost < minCost) {
            minCost = cost;
        }
    }
    
    return minCost;
}

int solution(int n, int m, int k) {
    return minCostToCut(n, k) + minCostToCut(m, k);
}

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

Time & Space Complexity

Time Complexity

⏱️

O(2^(n+m))

Each log can be cut in exponential ways, exploring all combinations

✓ Linear Growth

Space Complexity

O(n+m)

Recursion stack depth proportional to log lengths

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Cut Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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