Minimum Time to Complete All Tasks

• Sort the tasks array based on the end_i value in ascending order.
• Create a boolean array timeOn of size 2001 (based on constraints), initialized to false. This array will track at which time points the computer is on.
• Initialize a counter totalTime = 0.
• Iterate through each task [start, end, duration] in the sorted array.
• For the current task, count how many time slots are already on within its interval [start, end]. Let this be coveredDuration. This is done by iterating from start to end and checking timeOn.
• If coveredDuration < duration, the task needs needed = duration - coveredDuration more time slots.
• To satisfy this, iterate backwards from end down to start. For each time t in this range:
  • If timeOn[t] is false, set it to true, increment totalTime, and decrement needed.
  • Stop when needed becomes 0.
• After iterating through all tasks, totalTime will hold the minimum total time the computer is on.

The algorithm begins by sorting the tasks based on their end times in ascending order. This is a common heuristic in interval-based problems that helps in making greedy choices. We use a boolean array, let's call it timeOn, to keep track of which time points the computer is turned on. The size of this array will be determined by the maximum possible end time (2000 in this problem). We iterate through the sorted tasks one by one. For each task [start, end, duration], we first determine how much of its required duration is already covered by the time slots we've already decided to turn on. This is done by scanning the timeOn array from start to end and counting the true values. If the covered duration is less than the required duration, we need to turn on additional time slots. To make a greedy choice, we turn on the latest available time slots within the task's interval [start, end]. We iterate backwards from end to start. For each time t, if timeOn[t] is false, we set it to true and decrement the count of needed slots. We repeat this until the task's duration requirement is met. A counter variable keeps track of the total number of true values in timeOn, which is the final answer.

1import java.util.Arrays;2import java.util.Comparator;3 4class Solution {5    public int findMinimumTime(int[][] tasks) {6        // Sort tasks by their end times7        Arrays.sort(tasks, Comparator.comparingInt(a -> a[1]));8 9        // The maximum possible time is 2000. We use a 1-based index for time.10        boolean[] timeOn = new boolean[2001];11        int totalTime = 0;12 13        for (int[] task : tasks) {14            int start = task[0];15            int end = task[1];16            int duration = task[2];17 18            // 1. Check how much duration is already covered19            int covered = 0;20            for (int t = start; t <= end; t++) {21                if (timeOn[t]) {22                    covered++;23                }24            }25 26            // 2. If not enough, turn on more slots27            if (covered < duration) {28                int needed = duration - covered;29                // 3. Greedily choose the latest available slots30                for (int t = end; t >= start && needed > 0; t--) {31                    if (!timeOn[t]) {32                        timeOn[t] = true;33                        needed--;34                        totalTime++;35                    }36                }37            }38        }39        return totalTime;40    }41}

• Sort tasks by end_i in ascending order.
• Initialize a boolean[] timeOn for O(1) lookups and a FenwickTree ft of size M+1 for range queries.
• Initialize total_time = 0.
• For each task [start, end, duration] in sorted tasks:
  • Calculate covered = ft.query(end) - ft.query(start - 1).
  • Calculate needed = duration - covered.
  • If needed > 0:
    • Iterate t from end down to start:
      • If needed == 0, break the inner loop.
      • If !timeOn[t]:
        • Set timeOn[t] = true.
        • Update the Fenwick Tree: ft.update(t, 1).
        • Increment total_time.
        • Decrement needed.
• Return total_time.

The overall greedy algorithm remains unchanged: sort tasks by their end times and for each task, satisfy its duration by greedily picking the latest available time slots. We use two structures to track the state: a boolean[] timeOn array for O(1) lookup of whether a specific time slot is active, and a FenwickTree to quickly calculate the number of active slots in any given range [start, end]. After sorting tasks by end time, we iterate through them. For a task [start, end, duration], we first calculate covered slots using the Fenwick Tree: covered = fenwickTree.query(end) - fenwickTree.query(start - 1). This is an O(log M) operation. If covered < duration, we need to activate needed = duration - covered more slots. We iterate backwards from t = end down to start, using the timeOn array to find an inactive slot. When we find one, we activate it by setting timeOn[t] = true, updating the Fenwick Tree (fenwickTree.update(t, 1)), and decrementing needed. The total number of updates to the Fenwick Tree across all tasks is equal to the final answer, leading to a much better time complexity.

1// Conceptual code assuming FenwickTree class exists2import java.util.Arrays;3import java.util.Comparator;4 5class Solution {6    public int findMinimumTime(int[][] tasks) {7        Arrays.sort(tasks, Comparator.comparingInt(a -> a[1]));8 9        boolean[] timeOn = new boolean[2001];10        // A Fenwick Tree would be used here for optimization.11        // FenwickTree ft = new FenwickTree(2001);12        int totalTime = 0;13 14        for (int[] task : tasks) {15            int start = task[0];16            int end = task[1];17            int duration = task[2];18 19            // In an optimized solution, this part would use the Fenwick tree.20            // int covered = ft.query(end) - ft.query(start - 1);21            // For now, we simulate the logic which is bottleneck.22            int covered = 0;23            for(int i = start; i <= end; i++) {24                if(timeOn[i]) covered++;25            }26 27            if (covered < duration) {28                int needed = duration - covered;29                totalTime += needed;30                for (int t = end; t >= start && needed > 0; t--) {31                    if (!timeOn[t]) {32                        timeOn[t] = true;33                        // ft.update(t, 1); // Update would happen here34                        needed--;35                    }36                }37            }38        }39        return totalTime;40    }41}42// Assume FenwickTree class is implemented elsewhere43// class FenwickTree { ... }