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2204. Distance to a Cycle in Undirected Graph 🔒

中文文档

Description

You are given a positive integer n representing the number of nodes in a connected undirected graph containing exactly one cycle. The nodes are numbered from 0 to n - 1 (inclusive).

You are also given a 2D integer array edges, where edges[i] = [node1i, node2i] denotes that there is a bidirectional edge connecting node1i and node2i in the graph.

The distance between two nodes a and b is defined to be the minimum number of edges that are needed to go from a to b.

Return an integer array answer of size n, where answer[i] is the minimum distance between the ith node and any node in the cycle.

 

Example 1:

Input: n = 7, edges = [[1,2],[2,4],[4,3],[3,1],[0,1],[5,2],[6,5]]
Output: [1,0,0,0,0,1,2]
Explanation:
The nodes 1, 2, 3, and 4 form the cycle.
The distance from 0 to 1 is 1.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 3 is 0.
The distance from 4 to 4 is 0.
The distance from 5 to 2 is 1.
The distance from 6 to 2 is 2.

Example 2:

Input: n = 9, edges = [[0,1],[1,2],[0,2],[2,6],[6,7],[6,8],[0,3],[3,4],[3,5]]
Output: [0,0,0,1,2,2,1,2,2]
Explanation:
The nodes 0, 1, and 2 form the cycle.
The distance from 0 to 0 is 0.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 1 is 1.
The distance from 4 to 1 is 2.
The distance from 5 to 1 is 2.
The distance from 6 to 2 is 1.
The distance from 7 to 2 is 2.
The distance from 8 to 2 is 2.

 

Constraints:

  • 3 <= n <= 105
  • edges.length == n
  • edges[i].length == 2
  • 0 <= node1i, node2i <= n - 1
  • node1i != node2i
  • The graph is connected.
  • The graph has exactly one cycle.
  • There is at most one edge between any pair of vertices.

Solutions

Solution 1: Topological Sorting

We can first convert the edges in $edges$ into an adjacency list $g$, where $g[i]$ represents all adjacent nodes of node $i$, represented as a set.

Next, we delete nodes layer by layer from the outside to the inside until only a cycle remains. The specific method is as follows:

We first find all nodes with a degree of $1$ and remove these nodes from the graph. If after deletion, the degree of its adjacent node becomes $1$, then we add it to the queue $q$. During this process, we record all deleted nodes in order as $seq$; and we use an array $f$ to record the adjacent node of each node that is closer to the cycle, i.e., $f[i]$ represents the adjacent node of node $i$ that is closer to the cycle.

Finally, we initialize an answer array $ans$ of length $n$, where $ans[i]$ represents the minimum distance from node $i$ to any node in the cycle, initially $ans[i] = 0$. Then, we start traversing from the end of $seq$. For each node $i$, we can get the value of $ans[i]$ from its adjacent node $f[i]$, i.e., $ans[i] = ans[f[i]] + 1$.

After the traversal, return the answer array $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.

Similar problems:

Python3

class Solution:
    def distanceToCycle(self, n: int, edges: List[List[int]]) -> List[int]:
        g = defaultdict(set)
        for a, b in edges:
            g[a].add(b)
            g[b].add(a)
        q = deque(i for i in range(n) if len(g[i]) == 1)
        f = [0] * n
        seq = []
        while q:
            i = q.popleft()
            seq.append(i)
            for j in g[i]:
                g[j].remove(i)
                f[i] = j
                if len(g[j]) == 1:
                    q.append(j)
            g[i].clear()
        ans = [0] * n
        for i in seq[::-1]:
            ans[i] = ans[f[i]] + 1
        return ans

Java

class Solution {
    public int[] distanceToCycle(int n, int[][] edges) {
        Set<Integer>[] g = new Set[n];
        Arrays.setAll(g, k -> new HashSet<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (g[i].size() == 1) {
                q.offer(i);
            }
        }
        int[] f = new int[n];
        Deque<Integer> seq = new ArrayDeque<>();
        while (!q.isEmpty()) {
            int i = q.poll();
            seq.push(i);
            for (int j : g[i]) {
                g[j].remove(i);
                f[i] = j;
                if (g[j].size() == 1) {
                    q.offer(j);
                }
            }
        }
        int[] ans = new int[n];
        while (!seq.isEmpty()) {
            int i = seq.pop();
            ans[i] = ans[f[i]] + 1;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> distanceToCycle(int n, vector<vector<int>>& edges) {
        unordered_set<int> g[n];
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].insert(b);
            g[b].insert(a);
        }
        queue<int> q;
        for (int i = 0; i < n; ++i) {
            if (g[i].size() == 1) {
                q.push(i);
            }
        }
        int f[n];
        int seq[n];
        int k = 0;
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            seq[k++] = i;
            for (int j : g[i]) {
                g[j].erase(i);
                f[i] = j;
                if (g[j].size() == 1) {
                    q.push(j);
                }
            }
            g[i].clear();
        }
        vector<int> ans(n);
        for (; k; --k) {
            int i = seq[k - 1];
            ans[i] = ans[f[i]] + 1;
        }
        return ans;
    }
};

Go

func distanceToCycle(n int, edges [][]int) []int {
	g := make([]map[int]bool, n)
	for i := range g {
		g[i] = map[int]bool{}
	}
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a][b] = true
		g[b][a] = true
	}
	q := []int{}
	for i := 0; i < n; i++ {
		if len(g[i]) == 1 {
			q = append(q, i)
		}
	}
	f := make([]int, n)
	seq := []int{}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		seq = append(seq, i)
		for j := range g[i] {
			delete(g[j], i)
			f[i] = j
			if len(g[j]) == 1 {
				q = append(q, j)
			}
		}
		g[i] = map[int]bool{}
	}
	ans := make([]int, n)
	for k := len(seq) - 1; k >= 0; k-- {
		i := seq[k]
		ans[i] = ans[f[i]] + 1
	}
	return ans
}

TypeScript

function distanceToCycle(n: number, edges: number[][]): number[] {
    const g: Set<number>[] = new Array(n).fill(0).map(() => new Set<number>());
    for (const [a, b] of edges) {
        g[a].add(b);
        g[b].add(a);
    }
    const q: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (g[i].size === 1) {
            q.push(i);
        }
    }
    const f: number[] = Array(n).fill(0);
    const seq: number[] = [];
    while (q.length) {
        const i = q.pop()!;
        seq.push(i);
        for (const j of g[i]) {
            g[j].delete(i);
            f[i] = j;
            if (g[j].size === 1) {
                q.push(j);
            }
        }
        g[i].clear();
    }
    const ans: number[] = Array(n).fill(0);
    while (seq.length) {
        const i = seq.pop()!;
        ans[i] = ans[f[i]] + 1;
    }
    return ans;
}