There are n computers numbered from 0 to n-1 connected by ethernet cables connections forming a network where connections[i] = [a, b] represents a connection between computers a and b. Any computer can reach any other computer directly or indirectly through the network.

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’s not possible, return -1. 

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.

Example 4:

Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
Output: 0

Constraints:

• 1 <= n <= 10^5
• 1 <= connections.length <= min(n*(n-1)/2, 10^5)
• connections[i].length == 2
• 0 <= connections[i][0], connections[i][1] < n
• connections[i][0] != connections[i][1]
• There are no repeated connections.
• No two computers are connected by more than one cable.

Solution 1: Union-Find

Time complexity: O(V+E)
Space complexity: O(V)

// Author: Huahua
class Solution {
public:
  int makeConnected(int n, vector<vector<int>>& connections) {
    if (connections.size() < n - 1) return -1;
    vector<int> p(n);
    iota(begin(p), end(p), 0);
    
    function<int(int)> find = [&](int x) {
      return p[x] == x ? x : p[x] = find(p[x]);
    };
    
    for (const auto& c : connections)
      p[find(c[0])] = find(c[1]);    
    
    unordered_set<int> s;
    for (int i = 0; i < n; ++i)
      s.insert(find(i));
    
    return s.size() - 1;        
  }
};

Solution 2: DFS

Time complexity: O(V+E)
Space complexity: O(V+E)

// Author: Huahua
class Solution {
public:
  int makeConnected(int n, vector<vector<int>>& connections) {
    if (connections.size() < n - 1) return -1;
    vector<vector<int>> g(n);
    for (const auto& c : connections) {
      g[c[0]].push_back(c[1]);
      g[c[1]].push_back(c[0]);
    }
    vector<int> seen(n);
    int count = 0;
    function<void(int)> dfs = [&](int cur) {
      for (int nxt : g[cur])
        if (!seen[nxt]++) dfs(nxt);      
    };
    for (int i = 0; i < n; ++i)
      if (!seen[i]++ && ++count)
        dfs(i);        
    return count - 1;
  }
};

The post 花花酱 LeetCode 1319. Number of Operations to Make Network Connected appeared first on Huahua's Tech Road.

A critical connection is a connection that, if removed, will make some server unable to reach some other server.

Return all critical connections in the network in any order.

Example 1:

Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.

Constraints:

• 1 <= n <= 10^5
• n-1 <= connections.length <= 10^5
• connections[i][0] != connections[i][1]
• There are no repeated connections.

Solution: Tarjan

Time complexity: O(v+e)
Space complexity: O(v+e)

// Author: Huahua
class Solution {
public:
  vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
    vector<vector<int>> g(n);
    vector<int> ts(n, INT_MAX);
    vector<vector<int>> ans;
    int t = 0;
    
    function<int(int, int)> tarjan = [&](int i, int p) {
      int min_i = ts[i] = ++t;
      for (int j : g[i]) {
        if (ts[j] == INT_MAX) {
          int min_j = tarjan(j, i);
          min_i = min(min_i, min_j);
          if (ts[i] < min_j)
            ans.push_back({i, j});
        } else if (j != p) {
          min_i = min(min_i, ts[j]);
        }
      }
      return min_i;
    };
    
    for (const auto& e : connections) {
      g[e[0]].push_back(e[1]);
      g[e[1]].push_back(e[0]);
    }
    
    tarjan(0, -1);
    return ans;
  }
};

The post 花花酱 LeetCode 1192. Critical Connections in a Network appeared first on Huahua's Tech Road.