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

Thinking

A node is eventually safe if every walk from it ends at a terminal and never enters a cycle. Searching from every node independently retraces the same chains.

Reverse the edges so original out-degree becomes in-degree. Kahn deletion from in-degree \(0\) removes exactly the nodes that cannot reach a cycle, i.e. the safe nodes. One linear pass is enough for \(n\le 10^4\).

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

Thinking

The topological approach builds a reverse graph and in-degrees. A three-color DFS on the original graph is enough: gray means the current path has a cycle; black after the recursion means every walk from that node is safe.

No reverse graph is stored. The answer is the nodes whose DFS returns true, still in linear time.

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;
};

802. Find Eventual Safe States

Description

Solutions

Solution 1

Solution 2

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