1135. Connecting Cities With Minimum Cost

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1135. Connecting Cities With Minimum Cost

Description

There are n cities labeled from 1 to n. You are given the integer n and an array connections where connections[i] = [xi, yi, costi] indicates that the cost of connecting city xi and city yi (bidirectional connection) is costi.

Return the minimum cost to connect all the n cities such that there is at least one path between each pair of cities. If it is impossible to connect all the n cities, return -1,

The cost is the sum of the connections’ costs used.

Example 1:

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Input: n = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
Output: 6
Explanation: Choosing any 2 edges will connect all cities so we choose the minimum 2.

Example 2:

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Input: n = 4, connections = [[1,2,3],[3,4,4]]
Output: -1
Explanation: There is no way to connect all cities even if all edges are used.

Constraints:

  • 1 <= n <= 10^4
  • 1 <= connections.length <= 10^4
  • connections[i].length == 3
  • 1 <= xi, yi <= n
  • xi != yi
  • 0 <= costi <= 10^5

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int minimumCost(int n, vector<vector<int>>& connections) {
        // space: O(n)
        vector<bool> visited(n + 1, false);
        // space: O(n)
        vector<vector<pair<int, int>>> graph(n + 1);
        for (auto& c : connections) {
            int u = c[0], v = c[1], cost = c[2];
            graph[u].push_back({v, cost});
            graph[v].push_back({u, cost});
        }
        // 1: (2, 3)
        // 2: (1, 3)
        // 3: (4, 4)
        // 4: (3, 4)

        // pq:  {3, 2}
        // O(nlogn)
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        cut(1, pq, graph, visited);
        int numOfConnection = n, minCost = 0;
        while (!pq.empty() && numOfConnection > 1) {
            auto v = pq.top();
            pq.pop();
            int cost = v[0], u = v[1];
            if (visited[u]) {
                continue;
            }
            minCost += cost;
            cut(u, pq, graph, visited);
            numOfConnection--;
        }
        if (numOfConnection == 1) {
            return minCost;
        } else {
            return -1;
        }
    }

    void cut(int root, priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>>& pq, vector<vector<pair<int, int>>>& graph, vector<bool>& visited) {
        visited[root] = true;
        for (auto& [u, cost] : graph[root]) {
            if (!visited[u]) {
                pq.push({cost, u});
            }
        }
    }
};