Leetcode #2127: Maximum Employees to Be Invited to a Meeting
In this guide, we solve Leetcode #2127 Maximum Employees to Be Invited to a Meeting 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
A company is organizing a meeting and has a list of n employees, waiting to be invited. They have arranged for a large circular table, capable of seating any number of employees.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Graph, Topological Sort
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: favorite = [2,2,1,2]
Output: 3
Explanation:
The above figure shows how the company can invite employees 0, 1, and 2, and seat them at the round table.
All employees cannot be invited because employee 2 cannot sit beside employees 0, 1, and 3, simultaneously.
Note that the company can also invite employees 1, 2, and 3, and give them their desired seats.
The maximum number of employees that can be invited to the meeting is 3.
Python Solution
class Solution:
def maximumInvitations(self, favorite: List[int]) -> int:
def max_cycle(fa: List[int]) -> int:
n = len(fa)
vis = [False] * n
ans = 0
for i in range(n):
if vis[i]:
continue
cycle = []
j = i
while not vis[j]:
cycle.append(j)
vis[j] = True
j = fa[j]
for k, v in enumerate(cycle):
if v == j:
ans = max(ans, len(cycle) - k)
break
return ans
def topological_sort(fa: List[int]) -> int:
n = len(fa)
indeg = [0] * n
dist = [1] * n
for v in fa:
indeg[v] += 1
q = deque(i for i, v in enumerate(indeg) if v == 0)
while q:
i = q.popleft()
dist[fa[i]] = max(dist[fa[i]], dist[i] + 1)
indeg[fa[i]] -= 1
if indeg[fa[i]] == 0:
q.append(fa[i])
return sum(dist[i] for i, v in enumerate(fa) if i == fa[fa[i]])
return max(max_cycle(favorite), topological_sort(favorite))
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.