1135. Connecting Cities With Minimum Cost

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https://leetcode.com/problems/connecting-cities-with-minimum-cost/

https://github.com/openset/leetcode/tree/master/problems/connecting-cities-with-minimum-cost

There are N cities numbered from 1 to N.

You are given connections, where each connections[i] = [city1, city2, cost] represents the cost to connect city1 and city2 together.  (A connection is bidirectional: connecting city1 and city2 is the same as connecting city2 and city1.)

Return the minimum cost so that for every pair of cities, there exists a path of connections (possibly of length 1) that connects those two cities together.  The cost is the sum of the connection costs used. If the task is impossible, return -1.

 

Example 1:

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:

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.

 

Note:

  1. 1 <= N <= 10000
  2. 1 <= connections.length <= 10000
  3. 1 <= connections[i][0], connections[i][1] <= N
  4. 0 <= connections[i][2] <= 10^5
  5. connections[i][0] != connections[i][1]
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Intuition

Min cost => start with minimum cost, form connections with min cost always
If connection allowed => connect, N--, ans += cost

In the end if N == 1 => return ans, else return -1 => invalid

Check boundary condition in the start if (connections.length < N - 1) return -1

Sort by cost ascending
No need of PQ, since elements are accessed in order, no adding back
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