Asked in: ZOMATO
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All Testcases Passed ✔
Solution
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
// ... rest of solution available after purchaseExplanation
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To solve the problem of finding the minimum time to complete all tasks with dependencies where tasks can run in parallel as soon as their prerequisites are done, you need to think carefully about the nature of the dependencies and how tasks can be scheduled respecting those constraints.
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Step 1: Understand the Problem as a Directed Acyclic Graph (DAG)
- The tasks can be viewed as nodes in a graph.
- Dependencies represent directed edges from task a (prerequisite) to task b (dependent).
- The graph is a DAG because tasks with cyclic dependencies cannot be completed.
- The problem is essentially to find the minimum time needed to "topologically sort" the tasks while considering that multiple tasks can be executed simultaneously.
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Step 2: Concept of Levels or Layers in the DAG
- Since tasks take one unit of time each, and tasks can start only after all dependencies are complete, the tasks can be arranged in layers.
- The tasks at the first layer (level 1) have no prerequisites and can start immediately.
- Tasks at layer 2 depend only on tasks from layer 1, and so on.
- The minimum time to complete all tasks is equal to the number of layers in this layered structure.
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Step 3: Data Structures to Use
- Represent the graph using adjacency lists for efficient traversal.
- Maintain an array to track the in-degree (number of prerequisites) of each task.
- Tasks with in-degree 0 have no dependencies and can start immediately.
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Step 4: Kahn’s Algorithm for Topological Sorting with Level Tracking
- Initialize a queue and add all tasks with in-degree 0 (no prerequisites).
- These tasks form the first level (can be completed at time 1).
- Use an array or map to record the time at which each task can be completed.
- For each task removed from the queue:
- For each of its dependent tasks, reduce the in-degree by 1.
- If the in-degree of a dependent task becomes 0, it means all prerequisites are done.
- Assign the dependent task a completion time one unit later than the current task's completion time (since tasks take one unit of time).
- Add the dependent task to the queue.
- Repeat until the queue is empty.
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Step 5: Computing the Minimum Completion Time
- The minimum time to complete all tasks is the maximum completion time assigned to any task.
- Since tasks can run in parallel, tasks with the same earliest completion time belong to the same layer.
- Tracking the completion times ensures you know exactly when each task can finish.
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Step 6: Handling Edge Cases
- If the graph has cycles (detected if some tasks never reach in-degree 0), it is impossible to complete all tasks.
- The problem constraints imply no cycles, but you should be prepared to detect such cases.
- If there are tasks with no dependencies, they start at time 1.
- If no dependencies exist at all (M=0), all tasks can run in parallel and complete in one unit of time.
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Step 7: Complexity Considerations
- The approach visits each edge and node once, leading to O(N + M) time complexity.
- This is efficient given constraints where N and M can be up to 10^5.
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Step 8: Intuitive Example
- For tasks 0,1,2,3,4 with dependencies:
- 0 → 1, 0 → 3, 1 → 2, 3 → 4
- Tasks with no prerequisites: task 0 at time 1.
- Tasks depending on 0: tasks 1 and 3 at time 2.
- Tasks depending on 1 and 3: tasks 2 and 4 at time 3.
- Hence, total minimum time = 3.
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Step 9: Summary of Thought Process
- Model tasks and dependencies as a DAG.
- Use topological sorting with a queue, tracking in-degree of tasks.
- Assign completion times based on prerequisites.
- The maximum completion time among all tasks is the answer.
- This approach leverages BFS-like processing of layers in the dependency graph.
- Parallel execution means tasks in the same layer can be done simultaneously, so count only layers.
By conceptualizing the problem as a level-by-level topological sorting and carefully tracking task completion times, you can determine the minimum time needed to complete all dependent tasks efficiently and clearly.
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