Leetcode #2493: Divide Nodes Into the Maximum Number of Groups
In this guide, we solve Leetcode #2493 Divide Nodes Into the Maximum Number of Groups 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
You are given a positive integer n representing the number of nodes in an undirected graph. The nodes are labeled from 1 to n.
Quick Facts
- Difficulty: Hard
- 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 = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
Output: 4
Explanation: As shown in the image we:
- Add node 5 to the first group.
- Add node 1 to the second group.
- Add nodes 2 and 4 to the third group.
- Add nodes 3 and 6 to the fourth group.
We can see that every edge is satisfied.
It can be shown that that if we create a fifth group and move any node from the third or fourth group to it, at least on of the edges will not be satisfied.
Python Solution
class Solution:
def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
g = [[] for _ in range(n)]
for a, b in edges:
g[a - 1].append(b - 1)
g[b - 1].append(a - 1)
d = defaultdict(int)
for i in range(n):
q = deque([i])
dist = [0] * n
dist[i] = mx = 1
root = i
while q:
a = q.popleft()
root = min(root, a)
for b in g[a]:
if dist[b] == 0:
dist[b] = dist[a] + 1
mx = max(mx, dist[b])
q.append(b)
elif abs(dist[b] - dist[a]) != 1:
return -1
d[root] = max(d[root], mx)
return sum(d.values())
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.