1319.Number of Operations to Make Network Connected

tags: Medium,DFS,BFS,Graph

1319. Number of Operations to Make Network Connected

題目描述

There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [

$a_i$,

$b_i$] represents a connection between computers

$a_i$ and

$b_i$. Any computer can reach any other computer directly or indirectly through the network.

You are given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.

Return the minimum number of times you need to do this in order to make all the computers connected. If it is not possible, return -1.

範例

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]]
Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

Constraints:

• 1 <= n <= 10⁵
• 1 <= connections.length <= min(n * (n - 1) / 2, 10⁵)
• connections[i].length == 2
• 0 <=
  $a_i$,
  $b_i$ < n
• $a_i$ !=
  $b_i$
• There are no repeated connections.
• No two computers are connected by more than one cable.

解答

Python

class Solution:
    def makeConnected(self, n: int, connections: List[List[int]]) -> int:
        if len(connections) < n - 1:
            return -1

        graph = defaultdict(list)
        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)

        components = 0
        visited = set()

        def dfs(node):
            visited.add(node)
            for nei in graph[node]:
                if nei not in visited:
                    dfs(nei)

        for i in range(n):
            if i not in visited:
                dfs(i)
                components += 1

        return components - 1

Ron ChenThu, Mar 23, 2023

class Solution:
    def makeConnected(self, n: int, connections: List[List[int]]) -> int:
        if len(connections) < n - 1:
            return -1
        sets_tag = [i for i in range(n)]
        sets = {
            i: {i}
            for i in range(n)
        }
        count = 0
        for conn in connections:
            if sets_tag[conn[0]] == sets_tag[conn[1]]:
                continue
            sets[sets_tag[conn[0]]] |= sets[sets_tag[conn[1]]]
            replace_tag = sets_tag[conn[1]]
            for i in sets[sets_tag[conn[1]]]:
                sets_tag[i] = sets_tag[conn[0]]
            sets[replace_tag] = set()
        group_count = 0
        for s in sets.values():
            if len(s) > 0:
                group_count += 1
        return group_count - 1

gpwork4uFri, Mar 24, 2023

Javascript

function makeConnected(n, connections) {
  if (connections.length < n - 1) return -1;

  const graph = new Array(n).fill(0).map(() => []);
  for (const [v1, v2] of connections) {
    graph[v1].push(v2);
    graph[v2].push(v1);
  }

  const visited = new Array(n).fill(false);
  let count = 0;

  // 從每台電腦出發開始找
  for (let i = 0; i < n; i++) {
    if (visited[i]) continue;
    count++;
    const stack = [i];
    // 標記相連的電腦
    while (stack.length) {
      const node = stack.pop();
      if (visited[node]) continue;
      visited[node] = true;
      for (const vertex of graph[node]) {
        stack.push(vertex);
      }
    }
  }

  return count - 1;
}

MarsgoatThu, Mar 23, 2023

Reference

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