LeetCode-in-All

207. Course Schedule

Medium

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.

Example 1:

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

Output: true

Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

Example 2:

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

Output: false

Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= a_i, b_i < numCourses
All the pairs prerequisites[i] are unique.

Solution

typedef struct node {
    int V;
    struct node* next;
} node;

bool dfs(node** adjlist, bool* visited, bool* recStack, int current);

bool canFinish(int numCourses, int** prerequisites, int prerequisitesSize, int* prerequisitesColSize) {
    node** adjlist = malloc(sizeof(node*) * numCourses);
    for (int i = 0; i < numCourses; i++) {
        adjlist[i] = NULL;
    }
    for (int i = 0; i < prerequisitesSize; i++) {
        node* Node = malloc(sizeof(node));
        Node->V = prerequisites[i][0];
        Node->next = adjlist[prerequisites[i][1]];
        adjlist[prerequisites[i][1]] = Node;
    }

    bool* visited = calloc(sizeof(bool), numCourses);
    bool* recStack = calloc(sizeof(bool), numCourses);

    for (int V = 0; V < numCourses; V++) {
        if (dfs(adjlist, visited, recStack, V)) {
            return false;
        }
    }
    return true;
}

bool dfs(node** adjlist, bool* visited, bool* recStack, int current) {
    if (recStack[current]) {
        return true;
    }
    if (visited[current]) {
        return false;
    }
    visited[current] = true;
    recStack[current] = true;

    for (node* Node = adjlist[current]; Node; Node = Node->next) {
        if (dfs(adjlist, visited, recStack, Node->V)) {
            return true;
        }
    }

    recStack[current] = false;
    return false;
}

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