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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[k][j]);
                }
            }
        }
        vector<bool> ans;
        for (auto& q : queries) {
            ans.push_back(f[q[0]][q[1]]);
        }
        return ans;
    }
};

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func checkIfPrerequisite(n int, prerequisites [][]int, queries [][]int) (ans []bool) {
    f := make([][]bool, n)
    for i := range f {
        f[i] = make([]bool, n)
    }
    for _, p := range prerequisites {
        f[p[0]][p[1]] = true
    }
    for k := 0; k < n; k++ {
        for i := 0; i < n; i++ {
            for j := 0; j < n; j++ {
                f[i][j] = f[i][j] || (f[i][k] && f[k][j])
            }
        }
    }
    for _, q := range queries {
        ans = append(ans, f[q[0]][q[1]])
    }
    return
}

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function checkIfPrerequisite(n: number, prerequisites: number[][], queries: number[][]): boolean[] {
    const f = Array.from({ length: n }, () => Array(n).fill(false));
    prerequisites.forEach(([a, b]) => (f[a][b] = true));
    for (let k = 0; k < n; ++k) {
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < n; ++j) {
                f[i][j] ||= f[i][k] && f[k][j];
            }
        }
    }
    return queries.map(([a, b]) => f[a][b]);
}

Solution 2: Topological Sorting

Similar to Solution 1, we create a 2D array $f$, where $f[i][j]$ indicates whether node $i$ can reach node $j$. Additionally, we create an adjacency list $g$, where $g[i]$ represents all successor nodes of node $i$, and an array $indeg$, where $indeg[i]$ represents the in-degree of node $i$.

Next, we iterate through the prerequisites array $prerequisites$. For each item $[a, b]$ in it, we update the adjacency list $g$ and the in-degree array $indeg$.

Then, we use topological sorting to compute the reachability between all pairs of nodes.

We define a queue $q$, initially adding all nodes with an in-degree of $0$ to the queue. Then, we continuously perform the following operations: remove the front node $i$ from the queue, then iterate through all nodes $j$ in $g[i]$, setting $f[i][j]$ to $true$. Next, we enumerate node $h$, and if $f[h][i]$ is $true$, we also set $f[h][j]$ to $true$. After this, we decrease the in-degree of $j$ by $1$. If the in-degree of $j$ becomes $0$, we add $j$ to the queue.

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)]
        g = [[] for _ in range(n)]
        indeg = [0] * n
        for a, b in prerequisites:
            g[a].append(b)
            indeg[b] += 1
        q = deque(i for i, x in enumerate(indeg) if x == 0)
        while q:
            i = q.popleft()
            for j in g[i]:
                f[i][j] = True
                for h in range(n):
                    f[h][j] = f[h][j] or f[h][i]
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        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];
        List<Integer>[] g = new List[n];
        int[] indeg = new int[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var p : prerequisites) {
            g[p[0]].add(p[1]);
            ++indeg[p[1]];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        while (!q.isEmpty()) {
            int i = q.poll();
            for (int j : g[i]) {
                f[i][j] = true;
                for (int h = 0; h < n; ++h) {
                    f[h][j] |= f[h][i];
                }
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        List<Boolean> ans = new ArrayList<>();
        for (var qry : queries) {
            ans.add(f[qry[0]][qry[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));
        vector<int> g[n];
        vector<int> indeg(n);
        for (auto& p : prerequisites) {
            g[p[0]].push_back(p[1]);
            ++indeg[p[1]];
        }
        queue<int> q;
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
            }
        }
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            for (int j : g[i]) {
                f[i][j] = true;
                for (int h = 0; h < n; ++h) {
                    f[h][j] |= f[h][i];
                }
                if (--indeg[j] == 0) {
                    q.push(j);
                }
            }
        }
        vector<bool> ans;
        for (auto& qry : queries) {
            ans.push_back(f[qry[0]][qry[1]]);
        }
        return ans;
    }
};

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func checkIfPrerequisite(n int, prerequisites [][]int, queries [][]int) (ans []bool) {
    f := make([][]bool, n)
    for i := range f {
        f[i] = make([]bool, n)
    }
    g := make([][]int, n)
    indeg := make([]int, n)
    for _, p := range prerequisites {
        a, b := p[0], p[1]
        g[a] = append(g[a], b)
        indeg[b]++
    }
    q := []int{}
    for i, x := range indeg {
        if x == 0 {
            q = append(q, i)
        }
    }
    for len(q) > 0 {
        i := q[0]
        q = q[1:]
        for _, j := range g[i] {
            f[i][j] = true
            for h := 0; h < n; h++ {
                f[h][j] = f[h][j] || f[h][i]
            }
            indeg[j]--
            if indeg[j] == 0 {
                q = append(q, j)
            }
        }
    }
    for _, q := range queries {
        ans = append(ans, f[q[0]][q[1]])
    }
    return
}

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function checkIfPrerequisite(n: number, prerequisites: number[][], queries: number[][]): boolean[] {
    const f = Array.from({ length: n }, () => Array(n).fill(false));
    const g: number[][] = Array.from({ length: n }, () => []);
    const indeg: number[] = Array(n).fill(0);
    for (const [a, b] of prerequisites) {
        g[a].push(b);
        ++indeg[b];
    }
    const q: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (indeg[i] === 0) {
            q.push(i);
        }
    }
    while (q.length) {
        const i = q.shift()!;
        for (const j of g[i]) {
            f[i][j] = true;
            for (let h = 0; h < n; ++h) {
                f[h][j] ||= f[h][i];
            }
            if (--indeg[j] === 0) {
                q.push(j);
            }
        }
    }
    return queries.map(([a, b]) => f[a][b]);
}

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