Leetcode #1601: Maximum Number of Achievable Transfer Requests
In this guide, we solve Leetcode #1601 Maximum Number of Achievable Transfer Requests 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
We have n buildings numbered from 0 to n - 1. Each building has a number of employees.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Backtracking, Enumeration
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: n = 5, requests = [[0,1],[1,0],[0,1],[1,2],[2,0],[3,4]]
Output: 5
Explantion: Let's see the requests:
From building 0 we have employees x and y and both want to move to building 1.
From building 1 we have employees a and b and they want to move to buildings 2 and 0 respectively.
From building 2 we have employee z and they want to move to building 0.
From building 3 we have employee c and they want to move to building 4.
From building 4 we don't have any requests.
We can achieve the requests of users x and b by swapping their places.
We can achieve the requests of users y, a and z by swapping the places in the 3 buildings.
Python Solution
class Solution:
def maximumRequests(self, n: int, requests: List[List[int]]) -> int:
def check(mask: int) -> bool:
cnt = [0] * n
for i, (f, t) in enumerate(requests):
if mask >> i & 1:
cnt[f] -= 1
cnt[t] += 1
return all(v == 0 for v in cnt)
ans = 0
for mask in range(1 << len(requests)):
cnt = mask.bit_count()
if ans < cnt and check(mask):
ans = cnt
return ans
Complexity
The time complexity is , and the space complexity is , where and are the lengths of the room change request list and the number of rooms, respectively. The space complexity is , where and are the lengths of the room change request list and the number of rooms, respectively.
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.