A network company wants to connect n cities, labeled from 1 to n, using fiber-optic cables. They can only install cables along certain pre-approved routes, and each route has a cost.

You're given an integer n and a list of connections where connections[i] = [xᵢ, yᵢ, costᵢ], meaning a cable can be placed between city $x_{i}$ and city $y_{i}$ at the given $cost_i$ . All connections are bidirectional.

The goal is to connect all cities with the minimum total cost, using only the given routes.

The selected routes should not create cycles (no redundant cables).

Every city must be reachable from every other city.

Return the minimum total cost to connect all cities. If it's not possible, return -1.

Example 1:

Input: n = 4,
connections = [
  [1, 2, 10],
  [1, 3, 6],
  [1, 4, 5],
  [2, 4, 15],
  [3, 4, 4]
]
Output: 19
Explanation: The cheapest way to connect all cities is to select these routes:
City 3 ↔ City 4 → cost 4
City 1 ↔ City 4 → cost 5
City 1 ↔ City 2 → cost 10
Total = 4 + 5 + 10 = 19

Example 2:

Input: n = 5,
connections = [
  [1, 2, 3],
  [1, 3, 4],
  [4, 5, 2],
  [2, 4, 6],
  [3, 4, 5]
]

Output: 14 
Explanation: 
To connect all 5 cities with the minimum total cost, we choose the following 4 connections (since we need n - 1 = 4 edges for 5 nodes): 
City 1 ↔ City 2 → cost 3
City 1 ↔ City 3 → cost 4
City 3 ↔ City 4 → cost 5
City 4 ↔ City 5 → cost 2
Total cost = 3 + 4 + 5 + 2 = 14

These connections ensure:
- All cities are connected.
- No cycles are formed.
- The total cost is minimized.

Example 3:

Input: n = 4,
connections = [
  [1, 2, 5],
  [3, 4, 7]
]
Output: -1
Explanation: There are two disconnected groups:
Group 1: Cities 1 and 2
Group 2: Cities 3 and 4
There is no way to connect all cities, so the output is -1.

Constraints:

1 <= n <= 10⁴

1 <= connections.length <= 10⁴

connections[i].length == 3

0 <= cost_i <= 10⁵

1 <= x_i, y_i <= n
x_i != y_i

92. Minimum Cost to Connect Cities (Kruskal)