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The Factory II - C++ Solution with O(n * log(k * T)) Time Complexity


There are N workers in a factory. Each worker packs the same type of product, and for each of them we know the time it takes them to pack one product.

Find out the minimum amount of time the workers need to pack at least K products.

Example:

Input: times = [4, 7, 3, 6, 7, 1], K = 60
Output:30
Explanation:
1st worker packs 7 products in 4 * 7 = 28 units of time
2nd worker packs 4 products in 7 * 4 = 28 units of time
3rd worker packs 10 products in 3 * 10 = 30 units of time
4th worker packs 5 products in 6 * 5 = 30 units of time
5th worker packs 4 products in 7 * 4 = 28 units of time
6th worker packs 30 products in 1 * 30 = 30 units of time
We have 7 + 4 + 10 + 5 + 4 + 30 = 60 total products
in max(28, 28, 30, 30, 28, 30) = 30 units of time
and this is the minimum amount of time
since in 29 units of time we can pack only
7 + 4 + 9 + 4 + 4 + 29 = 57 products

Note:

Your algorithm should run in O(n * log(k * T)) time and use O(1) extra space.


Understanding the Problem

The core challenge of this problem is to determine the minimum time required for all workers to collectively pack at least K products. This problem is significant in scenarios where optimizing production time is crucial, such as in manufacturing and logistics.

Potential pitfalls include misunderstanding the distribution of work among workers and incorrectly calculating the total products packed within a given time frame.

Approach

To solve this problem, we can use a binary search approach on the time required. The idea is to determine the minimum time required such that the total number of products packed by all workers is at least K.

Naive Solution

A naive solution would involve iterating through each possible time unit and calculating the total products packed by all workers. This approach is not optimal due to its high time complexity.

Optimized Solution

We can optimize the solution using binary search. The steps are as follows:

  1. Initialize the search range for time from 1 to the maximum possible time (K * max_time_per_product).
  2. For each midpoint in the search range, calculate the total number of products packed by all workers within that time.
  3. If the total products packed is at least K, adjust the search range to find a smaller possible time.
  4. Otherwise, increase the search range to find a larger possible time.

Algorithm

Here is a step-by-step breakdown of the binary search algorithm:

  1. Initialize low to 1 and high to K * max_time_per_product.
  2. While low is less than or equal to high:
    1. Calculate mid as the average of low and high.
    2. Calculate the total number of products packed by all workers within mid time.
    3. If the total products packed is at least K, update result to mid and set high to mid - 1.
    4. Otherwise, set low to mid + 1.
  3. Return the result.

Code Implementation


#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

// Function to calculate the total products packed in given time
long long totalProductsPacked(const vector<int>& times, long long time) {
    long long total = 0;
    for (int t : times) {
        total += time / t;
    }
    return total;
}

// Function to find the minimum time to pack at least K products
long long minTimeToPackProducts(const vector<int>& times, int K) {
    long long low = 1;
    long long high = (long long) *max_element(times.begin(), times.end()) * K;
    long long result = high;

    while (low <= high) {
        long long mid = low + (high - low) / 2;
        if (totalProductsPacked(times, mid) >= K) {
            result = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }

    return result;
}

// Example usage
int main() {
    vector<int> times = {4, 7, 3, 6, 7, 1};
    int K = 60;
    long long result = minTimeToPackProducts(times, K);
    printf("Minimum time to pack %d products is %lld\n", K, result);
    return 0;
}

Complexity Analysis

The time complexity of the binary search approach is O(n * log(k * T)), where n is the number of workers, k is the number of products, and T is the maximum time per product. The space complexity is O(1) as we are using a constant amount of extra space.

Edge Cases

Potential edge cases include:

Each algorithm handles these edge cases by ensuring the binary search covers the entire possible range of times and correctly calculates the total products packed.

Testing

To test the solution comprehensively, consider the following test cases:

Use testing frameworks like Google Test for automated testing.

Thinking and Problem-Solving Tips

When approaching such problems:

Practice solving similar problems and study various algorithms to improve problem-solving skills.

Conclusion

In this blog post, we discussed the problem of finding the minimum time required for workers to pack at least K products. We explored a binary search approach to optimize the solution and provided a detailed explanation of the algorithm and its implementation in C++. Understanding and solving such problems is crucial for optimizing production processes in real-world scenarios.

Keep practicing and exploring further to enhance your problem-solving skills!

Additional Resources