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1971. Find if Path Exists in Graph

Description

There is a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1 (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself.

You want to determine if there is a valid path that exists from vertex source to vertex destination.

Given edges and the integers n, source, and destination, return true if there is a valid path from source to destination, or false otherwise.

 

Example 1:

Input: n = 3, edges = [[0,1],[1,2],[2,0]], source = 0, destination = 2
Output: true
Explanation: There are two paths from vertex 0 to vertex 2:
- 0 → 1 → 2
- 0 → 2

Example 2:

Input: n = 6, edges = [[0,1],[0,2],[3,5],[5,4],[4,3]], source = 0, destination = 5
Output: false
Explanation: There is no path from vertex 0 to vertex 5.

 

Constraints:

  • 1 <= n <= 2 * 105
  • 0 <= edges.length <= 2 * 105
  • edges[i].length == 2
  • 0 <= ui, vi <= n - 1
  • ui != vi
  • 0 <= source, destination <= n - 1
  • There are no duplicate edges.
  • There are no self edges.

Solutions

Solution 1: DFS

We first convert $\textit{edges}$ into an adjacency list $g$, then use DFS to determine whether there is a path from $\textit{source}$ to $\textit{destination}$.

During the process, we use an array $\textit{vis}$ to record the vertices that have already been visited to avoid revisiting them.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of nodes and edges, respectively.

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class Solution:
    def validPath(
        self, n: int, edges: List[List[int]], source: int, destination: int
    ) -> bool:
        def dfs(i: int) -> bool:
            if i == destination:
                return True
            vis.add(i)
            for j in g[i]:
                if j not in vis and dfs(j):
                    return True
            return False

        g = [[] for _ in range(n)]
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        vis = set()
        return dfs(source)

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class Solution {
    private int destination;
    private boolean[] vis;
    private List<Integer>[] g;

    public boolean validPath(int n, int[][] edges, int source, int destination) {
        this.destination = destination;
        vis = new boolean[n];
        g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        return dfs(source);
    }

    private boolean dfs(int i) {
        if (i == destination) {
            return true;
        }
        vis[i] = true;
        for (var j : g[i]) {
            if (!vis[j] && dfs(j)) {
                return true;
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool validPath(int n, vector<vector<int>>& edges, int source, int destination) {
        vector<int> g[n];
        vector<bool> vis(n);
        for (const auto& e : edges) {
            int u = e[0], v = e[1];
            g[u].push_back(v);
            g[v].push_back(u);
        }
        function<bool(int)> dfs = [&](int i) -> bool {
            if (i == destination) {
                return true;
            }
            vis[i] = true;
            for (int j : g[i]) {
                if (!vis