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
Your algorithm should run in O(n * log(k * T)) time and use O(1) extra space.
The core challenge of this problem is to determine the minimum time required for all workers to 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 relationship between time and the number of products packed by each worker, and not considering the maximum time taken by any worker as the limiting factor.
To solve this problem, we can use a binary search approach on the time. The idea is to find the minimum time required such that the total number of products packed by all workers is at least K.
1. **Naive Solution**: Iterate through each possible time and check if the workers can pack at least K products. This is not optimal as it can be very slow for large values of K and T.
2. **Optimized Solution**: Use binary search to find the minimum time. This approach is efficient and meets the required time complexity of O(n * log(k * T)).
1. Initialize the binary search range: `left` as 1 and `right` as the maximum possible time (K * max time in the array).
2. Perform binary search:
// Function to check if we can pack at least K products in given time
function canPackProductsInTime(times, K, mid) {
let totalProducts = 0;
for (let time of times) {
totalProducts += Math.floor(mid / time);
if (totalProducts >= K) {
return true;
}
}
return false;
}
// Main function to find the minimum time to pack at least K products
function minTimeToPackProducts(times, K) {
let left = 1;
let right = K * Math.max(...times);
let result = right;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (canPackProductsInTime(times, K, mid)) {
result = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
// Example usage
const times = [4, 7, 3, 6, 7, 1];
const K = 60;
console.log(minTimeToPackProducts(times, K)); // Output: 30
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 taken by any worker to pack one product. The space complexity is O(1) as we are using a constant amount of extra space.
Potential edge cases include:
Each algorithm handles these cases by ensuring the binary search covers the entire possible range of times and checks the total products packed correctly.
To test the solution comprehensively, consider a variety of test cases:
Using a testing framework like Jest can help automate and validate these test cases.
When approaching such problems:
Practice solving similar problems and study different algorithms to improve problem-solving skills.
In this blog post, we discussed how to solve the problem of finding the minimum time required for workers to pack at least K products using a binary search approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for optimizing production processes in real-world scenarios.
For further reading and practice, consider the following resources: