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] = [a_i, b_i] indicates that method a_i invokes method b_i.

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] == [a_i, b_i]
0 <= a_i, b_i <= n - 1
a_i != b_i
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.

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 = [&](auto&& dfs, int i) -> void {
            suspicious[i] = true;
            for (int j : g[i]) {
                if (!suspicious[j]) {
                    dfs(dfs, j);
                }
            }
        };
        dfs(dfs, k);
        auto dfs2 = [&](auto&& dfs2, int i) -> void {
            vis[i] = true;
            for (int j : f[i]) {
                if (!vis[j]) {
                    suspicious[j] = false;
                    dfs2(dfs2, j);
                }
            }
        };
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i] && !vis[i]) {
                dfs2(dfs2, i);
            }
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i]) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

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]]

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]);
}

3310 - Remove Methods From Project

3310. Remove Methods From Project

Description

Solutions

Solution 1: Two DFS

All Problems

All Solutions

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3310. Remove Methods From Project

Description

Solutions

Solution 1: Two DFS

All Problems

All Solutions