How do you solve Maximum Number of Achievable Transfer Requests on LeetCode?

Maximum Number of Achievable Transfer Requests (LeetCode #1601) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with undefined time complexity and undefined space complexity. Key insight: Instead of checking all combinations, we can use a backtracking approach to explore the requests and keep track of the net changes dynamically. This is like exploring paths in a maze, where we backtrack when we hit a dead end.

What is the time complexity of Maximum Number of Achievable Transfer Requests?

The optimal solution for Maximum Number of Achievable Transfer Requests runs in undefined time and uses undefined space. undefined

What is the brute force solution for Maximum Number of Achievable Transfer Requests?

The brute force approach for Maximum Number of Achievable Transfer Requests is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of requests to see which ones can be fulfilled while maintaining the balance of employees in each building. This is like checking every possible way to swap items in a box to see if you can get them all to the right places. The optimal Optimal Solution approach improves this to undefined.

What algorithm or data structure is used to solve Maximum Number of Achievable Transfer Requests?

Maximum Number of Achievable Transfer Requests uses the following concepts: Array, Backtracking, Bit Manipulation, Enumeration. The recommended patterns to study are: .

#1601

Maximum Number of Achievable Transfer Requests

Hard

ArrayBacktrackingBit ManipulationEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)
Space	O(1)

💡

Intuition

Time Space

Instead of checking all combinations, we can use a backtracking approach to explore the requests and keep track of the net changes dynamically. This is like exploring paths in a maze, where we backtrack when we hit a dead end.

⚙️

Algorithm

3 steps

1Step 1: Use a backtracking function to explore each request.
2Step 2: Maintain a count of requests that can be fulfilled.
3Step 3: For each request, either include it in the count or skip it and check the next request.

solution.py20 lines

1def maximumRequests(n, requests):
2    def backtrack(index, count, net_change):
3        nonlocal max_count
4        if index == len(requests):
5            if all(change == 0 for change in net_change):
6                max_count = max(max_count, count)
7            return
8        # Include the current request
9        fr, to = requests[index]
10        net_change[fr] -= 1
11        net_change[to] += 1
12        backtrack(index + 1, count + 1, net_change)
13        net_change[fr] += 1
14        net_change[to] -= 1
15        # Exclude the current request
16        backtrack(index + 1, count, net_change)
17
18    max_count = 0
19    backtrack(0, 0, [0] * n)
20    return max_count

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.