pairs, source, target):     graph = defaultdict(dict)     for s1, s2, rate_val1, rate_val2 in currency_pairs:           graph[s1][s2] = rate_val1           graph[s2][s1] = rate_val2       def backtrack(current, seen):           if current == target:                   return 1           product = 0

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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What is the time and space complexity of this function? It is a dfs function which goes through every possible path from node to another without cycles. Time complexity could be either O(n!) or O(2^n), or is there another answer? What is the space complexity?

def dfs(currency_pairs, source, target):
    graph = defaultdict(dict)
    for s1, s2, rate_val1, rate_val2 in currency_pairs:
          graph[s1][s2] = rate_val1
          graph[s2][s1] = rate_val2
 

    def backtrack(current, seen):
          if current == target:
                  return 1

          product = 0
          if currentingraph:
                neighbors = graph[current]
                for neighbor in neighbors:
                       if neighbor not in seen:
                            seen.add(neighbor)
                            product = max(product, graph[current][neighbor] * backtrack(neighbor, seen))
                            seen.remove(neighbor)
          return product
     return backtrack(source, {source})
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