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1462. Course Schedule IV

Description

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] = [ai, bi] indicates that you must take course ai first if you want to take course bi.

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

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.

Return a boolean array answer, where answer[j] is the answer to the jth query.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

 

Constraints:

  • 2 <= numCourses <= 100
  • 0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
  • prerequisites[i].length == 2
  • 0 <= ai, bi <= numCourses - 1
  • ai != bi
  • All the pairs [ai, bi] are unique.
  • The prerequisites graph has no cycles.
  • 1 <= queries.length <= 104
  • 0 <= ui, vi <= numCourses - 1
  • ui != vi

Solutions

Solution 1: Floyd's Algorithm

We create a 2D array $f$, where $f[i][j]$ indicates whether node $i$ can reach node $j$.

Next, we iterate through the prerequisites array $prerequisites$. For each item $[a, b]$ in it, we set $f[a][b]$ to $true$.

Then, we use Floyd's algorithm to compute the reachability between all pairs of nodes.

Specifically, we use three nested loops: first enumerating the intermediate node $k$, then the starting node $i$, and finally the ending node $j$. For each iteration, if node $i$ can reach node $k$ and node $k$ can reach node $j$, then node $i$ can also reach node $j$, and we set $f[i][j]$ to $true$.

After computing the reachability between all pairs of nodes, for each query $[a, b]$, we can directly return $f[a][b]$.

The time complexity is $O(n^3)$, and the space complexity is $O(n^2)$, where $n$ is the number of nodes.

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class Solution:
    def checkIfPrerequisite(
        self, n: int, prerequisites: List[List[int]], queries: List[List[int]]
    ) -> List[bool]:
        f = [[False] * n for _ in range(n)]
        for a, b in prerequisites:
            f[a][b] = True
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    if f[i][k] and f[k][j]:
                        f[i][j] = True
        return [f[a][b] for a, b in queries]

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class Solution {
    public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) {
        boolean[][] f = new boolean[n][n];
        for (var p : prerequisites) {
            f[p[0]][p[1]] = true;
        }
        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    f[i][j] |= f[i][k] && f[k][j];
                }
            }
        }
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            ans.add(f[q[0]][q[1]]);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<bool> checkIfPrerequisite(int n, vector<vector<int>>& prerequisites, vector<vector<int>>& queries) {
        bool f[n][n];
        memset(f, false, sizeof(f));
        for (auto& p : prerequisites) {
            f[p[0]][p[1]] = true;
        }
        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    f[i][j] |= (f[i][k] && f