Leetcode #1319: Number of Operations to Make Network Connected
In this guide, we solve Leetcode #1319 Number of Operations to Make Network Connected in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [ai, bi] represents a connection between computers ai and bi. Any computer can reach any other computer directly or indirectly through the network.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Graph
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
def find(x: int) -> int:
if p[x] != x:
p[x] = find(p[x])
return p[x]
cnt = 0
p = list(range(n))
for a, b in connections:
pa, pb = find(a), find(b)
if pa == pb:
cnt += 1
else:
p[pa] = pb
n -= 1
return -1 if n - 1 > cnt else n - 1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.