710142. Minimum time to fulfil all orders
Minimum time to fulfil all orders
Slug: minimum-time-to-fulfil-all-orders
Difficulty: Hard
Topic Tags: Binary Search, Algorithms
Company Tags: None
Summary
Given a list of tasks with their corresponding deadlines and a single machine with an available capacity, the problem is to find the minimum time required to complete all tasks. The tasks are represented as intervals on a timeline, and the goal is to schedule them in a way that minimizes the total time taken.
Detailed Explanation
The given task intervals can be thought of as a set of events that need to be scheduled. We can use binary search to find the minimum time required to complete all tasks. The idea is to start with a guess for the minimum time and then adjust it based on whether we can schedule all tasks within that time or not.
Here's a step-by-step breakdown of the solution:
- Initialize a variable
lowas 0, which represents the earliest possible deadline. - Initialize a variable
highas the maximum deadline, which represents the latest possible deadline. - Perform binary search to find the minimum time required to complete all tasks.
- Calculate the mid-time by averaging
lowandhigh. - Check if it's possible to schedule all tasks within the given time frame (i.e., if the sum of task durations is less than or equal to the mid-time).
- If yes, update
highas the mid-time. - If no, update
lowas the mid-time.
- If yes, update
- Calculate the mid-time by averaging
- Repeat step 3 until
highandlowconverge.
The time complexity of this algorithm is O(log N), where N is the number of tasks. The space complexity is O(1).
Optimized Solutions
Java
public int minTime(int[][] orders) {
Arrays.sort(orders, (a, b) -> a[0] - b[0]);
int low = 0;
int high = orders[orders.length - 1][1];
while (low < high) {
int mid = (low + high) / 2;
boolean canSchedule = true;
int currentTime = 0;
for (int[] order : orders) {
if (order[0] > mid) break;
if (currentTime >= order[1]) continue;
currentTime += order[0];
}
if (!canSchedule) high = mid;
else low = mid + 1;
}
return low;
}