What is the time complexity of Minimum Time to Complete All Tasks?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Minimum Time to Complete All Tasks?

This problem uses Array, Binary Search, Stack, Greedy, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Time to Complete All Tasks?

Insight 1: Sorting tasks by their end times is crucial for a greedy approach. It allows us to prioritize tasks that need to finish earlier, minimizing overlap and wasted computer time. Insight 2: Iterating backwards from the end time when filling the required duration guarantees the minimum overall time. Turning on the computer as late as possible within a task's valid time window helps avoid unnecessary overlap with subsequent tasks.

Which companies ask LeetCode 2589. Minimum Time to Complete All Tasks in interviews?

This problem is frequently asked at Snap.

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Minimum Time to Complete All Tasks - LeetCode 2589 Solution

Minimum Time to Complete All Tasks - Complete Solution Guide

Minimum Time to Complete All Tasks is LeetCode problem 2589, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a computer that can run an unlimited number of tasks at the same time . You are given a 2D integer array tasks where tasks[i] = [start i , end i , duration i ] indicates that the i th task should run for a total of duration i seconds (not necessarily continuous) within the inclusive time range [start i , end i ] . You may turn on the computer only when it needs to run a task. You can also turn it off if it is idle. Return the minimum time during which the computer should be turned on to

Detailed Explanation

The problem requires us to determine the minimum time the computer needs to be turned on to complete all given tasks. Each task has a start time, end time, and duration. The computer can run multiple tasks simultaneously. We must choose time slots within the task's start and end range such that the total running time for each task equals its duration. The goal is to minimize the total time the computer is on.

Solution Approach

The provided solution utilizes a greedy approach. The tasks are sorted by their end times. Then, for each task, the solution checks how much of its duration is already covered by the currently active time slots. If the duration isn't fully covered, it fills the remaining duration by activating time slots starting from the task's end time and working backwards. The algorithm ensures each task is completed within its time constraints while attempting to minimize overall time.

Step-by-Step Algorithm

Step 1: Sort the tasks array by their end times in ascending order. This step prioritizes tasks that need to finish sooner.
Step 2: Initialize a boolean array `time_on` (or similar) to represent whether each time slot is active (computer is on). The size of this array is based on the maximum possible end time (2000 in this case).
Step 3: Iterate through the sorted tasks.
Step 4: For each task, determine how much of its duration is already covered by checking the `time_on` array within the task's start and end times.
Step 5: Calculate the remaining duration that needs to be covered.
Step 6: If the remaining duration is greater than 0, iterate backwards from the task's end time to its start time.
Step 7: In this backward iteration, find unused time slots (where `time_on[t]` is false), activate them (`time_on[t] = true`), and decrement the remaining duration until it becomes 0.
Step 8: After processing all tasks, count the number of `true` values in the `time_on` array. This count represents the minimum time the computer needs to be on.

Key Insights

Insight 1: Sorting tasks by their end times is crucial for a greedy approach. It allows us to prioritize tasks that need to finish earlier, minimizing overlap and wasted computer time.
Insight 2: Iterating backwards from the end time when filling the required duration guarantees the minimum overall time. Turning on the computer as late as possible within a task's valid time window helps avoid unnecessary overlap with subsequent tasks.
Insight 3: Maintaining a boolean array representing whether each time slot is active allows efficient tracking of the time already used and avoids redundant calculations.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(m)

Topics

This problem involves: Array, Binary Search, Stack, Greedy, Sorting.

Companies

Asked at: Snap.

Minimum Time to Complete All Tasks | LeetCode 2589

Hard

Array Binary Search Stack Greedy Sorting

Time: O(n*m)Space: O(m)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int findMinimumTime(int[][] tasks) {
        Arrays.sort(tasks, (a, b) -> Integer.compare(a[1], b[1]));

        boolean[] timeOn = new boolean[2001];

        for (int[] task : tasks) {
            int start = task[0];
            int end = task[1];
            int duration = task[2];

            int alreadyCovered = 0;
            for (int t = start; t <= end; t++) {
                if (timeOn[t]) {
                    alreadyCovered++;
                }
            }

            int needed = duration - alreadyCovered;

            if (needed > 0) {
                for (int t = end; t >= start; t--) {
                    if (!timeOn[t]) {
                        timeOn[t] = true;
                        needed--;
                        if (needed == 0) {
                            break;
                        }
                    }
                }
            }
        }

        int sum = 0;
        for (boolean b : timeOn) {
            if (b) {
                sum++;
            }
        }
        return sum;
    }
}

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Problem Info

Description

Problem 2589: Minimum Time to Complete All Tasks - Detailed explanation and solution approaches.

Topics

Array Binary Search Stack Greedy Sorting

Companies

Snap

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2589

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Sort the tasks array by their end times in ascending order. This step prioritizes tasks that need to finish sooner.
2
Step 2: Initialize a boolean array `time_on` (or similar) to represent whether each time slot is active (computer is on). The size of this array is based on the maximum possible end time (2000 in this case).
3
Step 3: Iterate through the sorted tasks.
4
Step 4: For each task, determine how much of its duration is already covered by checking the `time_on` array within the task's start and end times.
5
Step 5: Calculate the remaining duration that needs to be covered.
6
Step 6: If the remaining duration is greater than 0, iterate backwards from the task's end time to its start time.
7
Step 7: In this backward iteration, find unused time slots (where `time_on[t]` is false), activate them (`time_on[t] = true`), and decrement the remaining duration until it becomes 0.
8
Step 8: After processing all tasks, count the number of `true` values in the `time_on` array. This count represents the minimum time the computer needs to be on.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(m)

📝

Why?

The time complexity is derived from the nested loops: iterating through each of the 'n' tasks, and for each task iterating a maximum of 'm' times to find the required time slots. Space complexity is for storing the timeOn array

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A dynamic programming approach could potentially be used, but it would likely be more complex and less efficient due to the potentially large range of time values and the need to consider all possible combinations of task scheduling.

Time: Potentially O(n*m^2) or worse, where n is number of tasks and m is the range of time.Space: O(m), depending on the DP state representation.

Trade-offs: Generally not a good choice for this problem due to increased complexity and likely worse performance compared to the greedy approach. Might be viable if the time range was extremely small and n was very large, though highly unlikely.

✨

Key Insights

1Insight 1: Sorting tasks by their end times is crucial for a greedy approach. It allows us to prioritize tasks that need to finish earlier, minimizing overlap and wasted computer time.
2Insight 2: Iterating backwards from the end time when filling the required duration guarantees the minimum overall time. Turning on the computer as late as possible within a task's valid time window helps avoid unnecessary overlap with subsequent tasks.
3Insight 3: Maintaining a boolean array representing whether each time slot is active allows efficient tracking of the time already used and avoids redundant calculations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Overlapping Tasks: The greedy approach efficiently handles overlapping tasks by ensuring the computer is turned on only when strictly necessary and only within the task's time range. Sorting ensures tasks with earlier deadlines are completed first.
Edge case 2: Tasks with Zero Duration (hypothetical, since duration >= 1): While not explicitly stated, tasks with duration 1 and start and end both the same are handled naturally by just turning the computer on at that single time.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting tasks by their end times. This can lead to a suboptimal solution where the computer is turned on for longer than necessary.

✗Common mistake 2: Turning on the computer from the beginning of the task's time range instead of the end. This can result in unnecessary overlap with future tasks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling in a data center. This algorithm could be adapted to minimize the usage time of a server while fulfilling a set of tasks with defined start and end times and required processing durations.

🏦Real-world scenario 2: Appointment scheduling. Consider a doctor scheduling appointments. Each appointment has a start time, end time, and duration. The algorithm can help minimize the total time the doctor spends working while satisfying all appointment requirements.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Time to Complete All Tasks"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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