Course Schedule

Course Schedule - Problem

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1.

You are given an array prerequisites where prerequisites[i] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

For example, the pair [0, 1] indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Input & Output

Example 1 — Basic Valid Case

$ Input: numCourses = 2, prerequisites = [[1,0]]

› Output: true

💡 Note: Course 1 requires course 0 first. Valid order: take course 0, then course 1. No circular dependency.

Example 2 — Circular Dependency

$ Input: numCourses = 2, prerequisites = [[1,0],[0,1]]

› Output: false

💡 Note: Course 1 requires course 0, but course 0 requires course 1. This creates a cycle, making it impossible to complete all courses.

Example 3 — No Prerequisites

$ Input: numCourses = 3, prerequisites = []

› Output: true

💡 Note: No prerequisites means all courses can be taken in any order. Always possible to complete.

Constraints

1 ≤ numCourses ≤ 10⁵
0 ≤ prerequisites.length ≤ 5000
prerequisites[i].length == 2
0 ≤ a_i, b_i < numCourses

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32 f Facebook 28

The key insight is that this is a cycle detection problem in a directed graph. If there's a cycle in the prerequisite dependencies, it's impossible to complete all courses. The optimal approach uses DFS with three states (unvisited, visiting, visited) to detect back edges indicating cycles. Time: O(V+E), Space: O(V+E).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

BFS Topological Sort (Kahn's Algorithm)

⏱️ Time: O(V + E) Space: O(V + E)

Calculate in-degrees for each course, start BFS from courses with zero in-degree. Remove edges and update in-degrees. If all courses are processed, return true.

Brute Force - Try All Orderings

⏱️ Time: O(n! × n²) Space: O(n)

Try every possible permutation of courses and check if each ordering respects all prerequisite constraints. This explores all n! possible orderings.

DFS Cycle Detection

⏱️ Time: O(V + E) Space: O(V + E)

Build a directed graph from prerequisites and use DFS with three states (unvisited, visiting, visited) to detect cycles. If a cycle exists, courses cannot be completed.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
    int delay, period;
    scanf("%d", &delay);
    scanf("%d", &period);
    printf("[\"customInterval\",\"customClearInterval\"]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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