3310. Remove Methods From Project

Description

You are maintaining a project that has n methods numbered from 0 to n - 1.

You are given two integers n and k, and a 2D integer array invocations, where invocations[i] = [ai, bi] indicates that method ai invokes method bi.

There is a known bug in method k. Method k, along with any method invoked by it, either directly or indirectly, are considered suspicious and we aim to remove them.

A group of methods can only be removed if no method outside the group invokes any methods within it.

Return an array containing all the remaining methods after removing all the suspicious methods. You may return the answer in any order. If it is not possible to remove all the suspicious methods, none should be removed.

 

Example 1:

Input: n = 4, k = 1, invocations = [[1,2],[0,1],[3,2]]

Output: [0,1,2,3]

Explanation:

Method 2 and method 1 are suspicious, but they are directly invoked by methods 3 and 0, which are not suspicious. We return all elements without removing anything.

Example 2:

Input: n = 5, k = 0, invocations = [[1,2],[0,2],[0,1],[3,4]]

Output: [3,4]

Explanation:

Methods 0, 1, and 2 are suspicious and they are not directly invoked by any other method. We can remove them.

Example 3:

Input: n = 3, k = 2, invocations = [[1,2],[0,1],[2,0]]

Output: []

Explanation:

All methods are suspicious. We can remove them.

 

Constraints:

• 1 <= n <= 105
• 0 <= k <= n - 1
• 0 <= invocations.length <= 2 * 105
• invocations[i] == [ai, bi]
• 0 <= ai, bi <= n - 1
• ai != bi
• invocations[i] != invocations[j]

Solutions

Solution 1: Two DFS

We can start from $k$ and find all suspicious methods, recording them in the array $\textit{suspicious}$. Then, we traverse from $0$ to $n-1$, starting from all non-suspicious methods, and mark all reachable methods as non-suspicious. Finally, we return all non-suspicious methods.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ represent the number of methods and the number of call relationships, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def remainingMethods(
        self, n: int, k: int, invocations: List[List[int]]
    ) -> List[int]:
        def dfs(i: int):
            suspicious[i] = True
            for j in g[i]:
                if not suspicious[j]:
                    dfs(j)

        def dfs2(i: int):
            vis[i] = True
            for j in f[i]:
                if not vis[j]:
                    suspicious[j] = False
                    dfs2(j)

        f = [[] for _ in range(n)]
        g = [[] for _ in range(n)]
        for a, b in invocations:
            f[a].append(b)
            f[b].append(a)
            g[a].append(b)
        suspicious = [False] * n
        dfs(k)

        vis = [False] * n
        ans = []
        for i in range(n):
            if not suspicious[i] and not vis[i]:
                dfs2(i)
        return [i for i in range(n) if not suspicious[i]]

class Solution {
    private boolean[] suspicious;
    private boolean[] vis;
    private List<Integer>[] f;
    private List<Integer>[] g;

    public List<Integer> remainingMethods(int n, int k, int[][] invocations) {
        suspicious = new boolean[n];
        vis = new boolean[n];
        f = new List[n];
        g = new List[n];
        Arrays.setAll(f, i -> new ArrayList<>());
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : invocations) {
            int a = e[0], b = e[1];
            f[a].add(b);
            f[b].add(a);
            g[a].add(b);
        }
        dfs(k);
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i] && !vis[i]) {
                dfs2(i);
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i]) {
                ans.add(i);
            }
        }
        return ans;
    }

    private void dfs(int i) {
        suspicious[i] = true;
        for (int j : g[i]) {
            if (!suspicious[j]) {
                dfs(j);
            }
        }
    }

    private void dfs2(int i) {
        vis[i] = true;
        for (int j : f[i]) {
            if (!vis[j]) {
                suspicious[j] = false;
                dfs2(j);
            }
        }
    }
}

class Solution {
public:
    vector<int> remainingMethods(int n, int k, vector<vector<int>>& invocations) {
        vector<bool> suspicious(n);
        vector<bool> vis(n);
        vector<int> f[n];
        vector<int> g[n];
        for (const auto& e : invocations) {
            int a = e[0], b = e[1];
            f[a].push_back(b);
            f[b].push_back(a);
            g[a].push_back(b);
        }
        auto dfs = [&](this auto&& dfs, int i) -> void {
            suspicious[i] = true;
            for (int j : g[i]) {
                if (!suspicious[j]) {
                    dfs(j);
                }
            }
        };
        dfs(k);
        auto dfs2 = [&](this auto&& dfs2, int i) -> void {
            vis[i] = true;
            for (int j : f[i]) {
                if (!vis[j]) {
                    suspicious[j] = false;
                    dfs2(j);
                }
            }
        };
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i] && !vis[i]) {
                dfs2(i);
            }
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i]) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

func remainingMethods(n int, k int, invocations [][]int) []int {
    suspicious := make([]bool, n)
    vis := make([]bool, n)
    f := make([][]int, n)
    g := make([][]int, n)

    for _, e := range invocations {
        a, b := e[0], e[1]
        f[a] = append(f[a], b)
        f[b] = append(f[b], a)
        g[a] = append(g[a], b)
    }

    var dfs func(int)
    dfs = func(i int) {
        suspicious[i] = true
        for _, j := range g[i] {
            if !suspicious[j] {
                dfs(j)
            }
        }
    }

    dfs(k)

    var dfs2 func(int)
    dfs2 = func(i int) {
        vis[i] = true
        for _, j := range f[i] {
            if !vis[j] {
                suspicious[j] = false
                dfs2(j)
            }
        }
    }

    for i := 0; i < n; i++ {
        if !suspicious[i] && !vis[i] {
            dfs2(i)
        }
    }

    var ans []int
    for i := 0; i < n; i++ {
        if !suspicious[i] {
            ans = append(ans, i)
        }
    }

    return ans
}

function remainingMethods(n: number, k: number, invocations: number[][]): number[] {
    const suspicious: boolean[] = Array(n).fill(false);
    const vis: boolean[] = Array(n).fill(false);
    const f: number[][] = Array.from({ length: n }, () => []);
    const g: number[][] = Array.from({ length: n }, () => []);

    for (const [a, b] of invocations) {
        f[a].push(b);
        f[b].push(a);
        g[a].push(b);
    }

    const dfs = (i: number) => {
        suspicious[i] = true;
        for (const j of g[i]) {
            if (!suspicious[j]) {
                dfs(j);
            }
        }
    };

    dfs(k);

    const dfs2 = (i: number) => {
        vis[i] = true;
        for (const j of f[i]) {
            if (!vis[j]) {
                suspicious[j] = false;
                dfs2(j);
            }
        }
    };

    for (let i = 0; i < n; i++) {
        if (!suspicious[i] && !vis[i]) {
            dfs2(i);
        }
    }

    return Array.from({ length: n }, (_, i) => i).filter(i => !suspicious[i]);
}

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