802. Find Eventual Safe States

Description

There is a directed graph of n nodes with each node labeled from 0 to n - 1. The graph is represented by a 0-indexed 2D integer array graph where graph[i] is an integer array of nodes adjacent to node i, meaning there is an edge from node i to each node in graph[i].

A node is a terminal node if there are no outgoing edges. A node is a safe node if every possible path starting from that node leads to a terminal node (or another safe node).

Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.

 

Example 1:

Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output: [2,4,5,6]
Explanation: The given graph is shown above.
Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them.
Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.

Example 2:

Input: graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]
Output: [4]
Explanation:
Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.

 

Constraints:

• n == graph.length
• 1 <= n <= 104
• 0 <= graph[i].length <= n
• 0 <= graph[i][j] <= n - 1
• graph[i] is sorted in a strictly increasing order.
• The graph may contain self-loops.
• The number of edges in the graph will be in the range [1, 4 * 104].

Solutions

Solution 1

Python3JavaC++GoJavaScript

class Solution:
    def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]:
        rg = defaultdict(list)
        indeg = [0] * len(graph)
        for i, vs in enumerate(graph):
            for j in vs:
                rg[j].append(i)
            indeg[i] = len(vs)
        q = deque([i for i, v in enumerate(indeg) if v == 0])
        while q:
            i = q.popleft()
            for j in rg[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    q.append(j)
        return [i for i, v in enumerate(indeg) if v == 0]

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        int[] indeg = new int[n];
        List<Integer>[] rg = new List[n];
        Arrays.setAll(rg, k -> new ArrayList<>());
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            for (int j : graph[i]) {
                rg[j].add(i);
            }
            indeg[i] = graph[i].length;
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        while (!q.isEmpty()) {
            int i = q.pollFirst();
            for (int j : rg[i]) {
                if (--indeg[j] == 0) {
                    q.offer(j);
                }
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                ans.add(i);
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<int> indeg(n);
        vector<vector<int>> rg(n);
        queue<int> q;
        for (int i = 0; i < n; ++i) {
            for (int j : graph[i]) rg[j].push_back(i);
            indeg[i] = graph[i].size();
            if (indeg[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            for (int j : rg[i])
                if (--indeg[j] == 0) q.push(j);
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i)
            if (indeg[i] == 0) ans.push_back(i);
        return ans;
    }
};

func eventualSafeNodes(graph [][]int) []int {
    n := len(graph)
    indeg := make([]int, n)
    rg := make([][]int, n)
    q := []int{}
    for i, vs := range graph {
        for _, j := range vs {
            rg[j] = append(rg[j], i)
        }
        indeg[i] = len(vs)
        if indeg[i] == 0 {
            q = append(q, i)
        }
    }
    for len(q) > 0 {
        i := q[0]
        q = q[1:]
        for _, j := range rg[i] {
            indeg[j]--
            if indeg[j] == 0 {
                q = append(q, j)
            }
        }
    }
    ans := []int{}
    for i, v := range indeg {
        if v == 0 {
            ans = append(ans, i)
        }
    }
    return ans
}

/**
 * @param {number[][]} graph
 * @return {number[]}
 */
var eventualSafeNodes = function (graph) {
    const n = graph.length;
    const rg = new Array(n).fill(0).map(() => new Array());
    const indeg = new Array(n).fill(0);
    const q = [];
    for (let i = 0; i < n; ++i) {
        for (let j of graph[i]) {
            rg[j].push(i);
        }
        indeg[i] = graph[i].length;
        if (indeg[i] == 0) {
            q.push(i);
        }
    }
    while (q.length) {
        const i = q.shift();
        for (let j of rg[i]) {
            if (--indeg[j] == 0) {
                q.push(j);
            }
        }
    }
    let ans = [];
    for (let i = 0; i < n; ++i) {
        if (indeg[i] == 0) {
            ans.push(i);
        }
    }
    return ans;
};

Solution 2

Python3JavaC++GoJavaScript

class Solution:
    def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]:
        def dfs(i):
            if color[i]:
                return color[i] == 2
            color[i] = 1
            for j in graph[i]:
                if not dfs(j):
                    return False
            color[i] = 2
            return True

        n = len(graph)
        color = [0] * n
        return [i for i in range(n) if dfs(i)]

class Solution {
    private int[] color;
    private int[][] g;

    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        color = new int[n];
        g = graph;
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (dfs(i)) {
                ans.add(i);
            }
        }
        return ans;
    }

    private boolean dfs(int i) {
        if (color[i] > 0) {
            return color[i] == 2;
        }
        color[i] = 1;
        for (int j : g[i]) {
            if (!dfs(j)) {
                return false;
            }
        }
        color[i] = 2;
        return true;
    }
}

class Solution {
public:
    vector<int> color;

    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        color.assign(n, 0);
        vector<int> ans;
        for (int i = 0; i < n; ++i)
            if (dfs(i, graph)) ans.push_back(i);
        return ans;
    }

    bool dfs(int i, vector<vector<int>>& g) {
        if (color[i]) return color[i] == 2;
        color[i] = 1;
        for (int j : g[i])
            if (!dfs(j, g)) return false;
        color[i] = 2;
        return true;
    }
};

func eventualSafeNodes(graph [][]int) []int {
    n := len(graph)
    color := make([]int, n)
    var dfs func(int) bool
    dfs = func(i int) bool {
        if color[i] > 0 {
            return color[i] == 2
        }
        color[i] = 1
        for _, j := range graph[i] {
            if !dfs(j) {
                return false
            }
        }
        color[i] = 2
        return true
    }
    ans := []int{}
    for i := range graph {
        if dfs(i) {
            ans = append(ans, i)
        }
    }
    return ans
}

/**
 * @param {number[][]} graph
 * @return {number[]}
 */
var eventualSafeNodes = function (graph) {
    const n = graph.length;
    const color = new Array(n).fill(0);
    function dfs(i) {
        if (color[i]) {
            return color[i] == 2;
        }
        color[i] = 1;
        for (const j of graph[i]) {
            if (!dfs(j)) {
                return false;
            }
        }
        color[i] = 2;
        return true;
    }
    let ans = [];
    for (let i = 0; i < n; ++i) {
        if (dfs(i)) {
            ans.push(i);
        }
    }
    return ans;
};

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