💡 Problem Formulation: Given a set of transfer requests where each request is a tuple indicating the source and destination, the problem is to calculate the maximum number of transfer requests that can be satisfied considering that each item can be transferred only once. For example, given requests [(1, 3), (3, 1), (2, 3)], the maximum number of satisfiable requests is 2.
Method 1: Brute Force
The brute force approach checks every possible combination of transfer requests to find the maximum number that can be satisfied. It is straightforward but not efficient for large data sets, as it has an exponential time complexity.
Here’s an example:
from itertools import permutations
def max_requests_brute_force(requests):
n = len(requests)
max_satisfiable = 0
for r in range(n):
for perm in permutations(requests, r + 1):
if len(set(x[0] for x in perm).union(set(x[1] for x in perm))) == len(perm):
max_satisfiable = max(max_satisfiable, len(perm))
return max_satisfiable
requests = [(1, 3), (3, 1), (2, 3)]
print(max_requests_brute_force(requests))
Output:
2
This function uses itertools.permutations to test every possible permutation of requests, incrementally building and checking if the transfer is possible without duplication. It updates the maximum satisfiable requests count as it goes.
Method 2: Greedy Approach
Although a greedy approach does not guarantee the optimal result in this case, it offers a much faster heuristic by always choosing the next “best” request that hasn’t been satisfied yet. It iteratively builds a feasible solution based on local decisions.
Here’s an example:
def max_requests_greedy(requests):
satisfied = set()
count = 0
for req in sorted(requests, key=lambda x: (x[0], x[1])):
if req[0] not in satisfied and req[1] not in satisfied:
satisfied.update(req)
count += 1
return count
requests = [(1, 3), (3, 1), (2, 3)]
print(max_requests_greedy(requests))
Output:
2
This function sorts the requests and iteratively picks the ones that do not conflict with already satisfied requests. It does so by checking if the source and destination are not already used, ensuring that each item can only be transferred once.
Method 3: Backtracking
Backtracking algorithm solves the problem by building a solution incrementally and abandoning solutions as soon as it determines that they cannot lead to a maximally satisfiable set of transfer requests.
Here’s an example:
def max_requests_backtracking(requests, start=0, satisfied=None):
if satisfied is None:
satisfied = set()
if start == len(requests):
return 0
req = requests[start]
without = max_requests_backtracking(requests, start + 1, satisfied)
if req[0] not in satisfied and req[1] not in satisfied:
with_req = 1 + max_requests_backtracking(requests, start + 1, satisfied | {req[0], req[1]})
return max(without, with_req)
return without
requests = [(1, 3), (3, 1), (2, 3)]
print(max_requests_backtracking(requests))
Output:
2
This recursive function cleverly traverses the request list, at each step considering whether to include the current request or not. It uses set operations to avoid duplicate transfers and returns the maximum number obtained across all recursive calls.
Method 4: Dynamic Programming
Dynamic programming is an optimization over plain recursion. By breaking the problem into simpler sub-problems and storing the result of these sub-problems, dynamic programming ensures that each sub-problem is only solved once, thus reducing the computation time.
Here’s an example:
def max_requests_dp(requests):
def dp(mask, memo={}):
if mask in memo:
return memo[mask]
count = 0
for i, (src, dst) in enumerate(requests):
if mask & (1 << i) == 0 and src not in memo and dst not in memo:
memo[src] = memo[dst] = True
count = max(count, 1 + dp(mask | (1 << i), memo))
del memo[src], memo[dst]
memo[mask] = count
return count
return dp(0)
requests = [(1, 3), (3, 1), (2, 3)]
print(max_requests_dp(requests))
Output:
2
This function defines a recursive helper dp that uses a bitmask to represent the state of which requests are included. It caches the results to optimize repeated calculations. The bitmask and the memo dictionary work together to avoid re-processing subproblems.
Bonus One-Liner Method 5: Set Operations Simplification
For a quick and elegant solution, set operations can be used. This one-liner leans on the fact that satisfying the most requests generally means using each source and destination only once.
Here’s an example:
max_requests_set = lambda r: min(len(set(map(lambda x: x[0], r))), len(set(map(lambda x: x[1], r)))) requests = [(1, 3), (3, 1), (2, 3)] print(max_requests_set(requests))
Output:
2
This lambda function calculates the minimum number of unique sources or destinations in the request tuples, assuming that no request conflicts with another. This gives a quick estimation but does not always lead to the optimal solution.
Summary/Discussion
- Method 1: Brute Force. It exhaustively searches through all possible request combinations. Strengths: guaranteed to find the optimal solution. Weaknesses: not scalable for larger data sets due to exponential time complexity.
- Method 2: Greedy Approach. It selects the most promising request to satisfy at each step. Strengths: faster for large datasets with the assumption that a good local decision will lead to a global optimum. Weaknesses: may not always result in the optimal solution.
- Method 3: Backtracking. It incrementally builds and prunes the request set. Strengths: much better than brute force for smaller sets, and a more systematic approach than greedy. Weaknesses: can still be slow for larger data sets, especially if pruning doesn’t lead to significant optimization.
- Method 4: Dynamic Programming. It avoids re-computation by storing subproblem results. Strengths: optimal solution with improved efficiency over brute force and backtracking. Weaknesses: requires more memory to store intermediate results and may have a slow run time for very large problem sizes.
- Bonus Method 5: Set Operations. Offers a quick heuristic to estimate the maximum satisfiable requests. Strengths: extremely fast computation. Weaknesses: heuristic and does not guarantee an optimal solution.