Maximum Number of Achievable Transfer Requests

We have n buildings numbered from 0 to n - 1. Each building has a number of employees. It's transfer season, and some employees want to change the building they reside in.

You are given an array requests where requests[i] = [from_i, to_i] represents an employee's request to transfer from building from_i to building to_i.

All buildings are full, so a list of requests is achievable only if for each building, the net change in employee transfers is zero. This means the number of employees leaving is equal to the number of employees moving in. For example if n = 3 and two employees are leaving building 0, one is leaving building 1, and one is leaving building 2, there should be two employees moving to building 0, one employee moving to building 1, and one employee moving to building 2.

Return the maximum number of achievable requests.

Example 1:

Input: n = 5, requests = [[0,1],[1,0],[0,1],[1,2],[2,0],[3,4]]
Output: 5

Example 2:

Input: n = 3, requests = [[0,0],[1,2],[2,1]]
Output: 3

Example 3:

Input: n = 4, requests = [[0,3],[3,1],[1,2],[2,0]]
Output: 4

What is the most efficient algorithm to solve this problem?

This problem asks us to find the maximum number of achievable employee transfer requests, given that the net change in employees for each building must be zero. We can solve this using a brute-force approach by trying all possible combinations of requests and checking if they satisfy the given condition. Since the number of requests is limited to 16, we can use bit manipulation to represent each combination.

Naive Approach:

1. Generate all possible subsets of the requests.
2. For each subset, check if it is achievable. A subset is achievable if for each building, the number of employees leaving is equal to the number of employees moving in.
3. Keep track of the maximum number of requests in an achievable subset.

def maximum_achievable_requests_naive(n, requests):
    max_requests = 0
    for i in range(1 << len(requests)):
        subset = []
        for j in range(len(requests)):
            if (i >> j) & 1:
                subset.append(requests[j])

        balance = [0] * n
        achievable = True
        for request in subset:
            balance[request[0]] -= 1
            balance[request[1]] += 1

        for b in balance:
            if b != 0:
                achievable = False
                break

        if achievable:
            max_requests = max(max_requests, len(subset))

    return max_requests

Optimal Approach (Bit Manipulation with Backtracking):

We can use a more efficient approach by iterating through all possible combinations of requests using bit manipulation. For each combination, we check if it's a valid arrangement by verifying if the net change in employees is zero for each building. This involves creating a balance array to track the change in employee count for each building.

def maximum_achievable_requests(n, requests):
    max_achievable = 0

    def check_achievability(mask):
        balance = [0] * n
        count = 0
        for i in range(len(requests)):
            if (mask >> i) & 1:
                balance[requests[i][0]] -= 1
                balance[requests[i][1]] += 1
                count += 1

        for b in balance:
            if b != 0:
                return 0
        return count

    for mask in range(1 << len(requests)):
        max_achievable = max(max_achievable, check_achievability(mask))

    return max_achievable

Big O Analysis:

• Time Complexity: O(2^R * N), where R is the number of requests and N is the number of buildings. We iterate through all possible combinations of requests (2^R), and for each combination, we check the balance for each building (N).
• Space Complexity: O(N), where N is the number of buildings. We use an array of size N to store the balance of employees for each building.

Edge Cases:

1. Empty Requests Array: If the requests array is empty, the function should return 0, as no requests can be achieved.
2. Requests within the same building: Requests where from_i == to_i are valid and should be included if they don't violate the net change constraint.
3. Unachievable Requests: If no combination of requests satisfies the condition (net change is zero for all buildings), the function should return 0.
4. Large Number of Buildings/Requests: The problem states that n <= 20 and requests.length <= 16. The exponential time complexity means this solution is only suitable for relatively small inputs. For larger inputs, a different approach might be necessary (possibly involving more complex graph algorithms or optimization techniques).

Example Usage:

n = 5
requests = [[0, 1], [1, 0], [0, 1], [1, 2], [2, 0], [3, 4]]
result = maximum_achievable_requests(n, requests)
print(f"Maximum achievable requests: {result}")  # Output: 5

n = 3
requests = [[0, 0], [1, 2], [2, 1]]
result = maximum_achievable_requests(n, requests)
print(f"Maximum achievable requests: {result}")  # Output: 3

n = 4
requests = [[0, 3], [3, 1], [1, 2], [2, 0]]
result = maximum_achievable_requests(n, requests)
print(f"Maximum achievable requests: {result}")  # Output: 4