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Maximize Amount After Two Days of Conversions

Difficulty: Medium

Problem Description

You are given a string initialCurrency, and you start with 1.0 of initialCurrency. You are also given four arrays with currency pairs (strings) and rates (real numbers): pairs1, rates1, pairs2, and rates2. You can perform any number of conversions on day 1, followed by any number of conversions on day 2. Return the maximum amount of initialCurrency you can have after performing any conversions on both days in order.

Key Insights

  • You can perform multiple conversions in a day.
  • Each day has its own set of conversion pairs and rates.
  • You can also convert back to the original currency at an inverse rate.
  • The goal is to maximize the final amount of the initial currency after two days of conversions.

Space and Time Complexity

Time Complexity: O(n^2 + m^2), where n and m are the lengths of pairs1 and pairs2 respectively, as we might need to explore all pairs for conversions. Space Complexity: O(n + m), for storing the conversion rates and results.

Solution

To solve the problem, we can use a graph-based approach. Each currency conversion can be represented as a directed edge in a graph where nodes are currencies and edges are conversion rates.

  1. Build a graph for the conversions on day 1 using pairs1 and rates1.
  2. Use a breadth-first search (BFS) or depth-first search (DFS) to explore all possible conversions from initialCurrency on day 1 to find the maximum amount of each currency you can achieve.
  3. Repeat the process for day 2 using pairs2 and rates2, starting from the amounts obtained from day 1.
  4. Keep track of the maximum amount of initialCurrency throughout the conversions across both days.

Code Solutions

def maximizeAmount(initialCurrency, pairs1, rates1, pairs2, rates2):
    from collections import defaultdict
    
    def getMaxAmounts(pairs, rates, startCurrency):
        graph = defaultdict(list)
        for (start, target), rate in zip(pairs, rates):
            graph[start].append((target, rate))
        
        maxAmounts = {startCurrency: 1.0}
        queue = [startCurrency]
        
        while queue:
            currCurrency = queue.pop(0)
            currAmount = maxAmounts[currCurrency]
            for target, rate in graph[currCurrency]:
                newAmount = currAmount * rate
                if target not in maxAmounts or newAmount > maxAmounts[target]:
                    maxAmounts[target] = newAmount
                    queue.append(target)
        
        return maxAmounts

    # Day 1 conversions
    day1Amounts = getMaxAmounts(pairs1, rates1, initialCurrency)
    
    # Day 2 conversions
    maxFinalAmount = day1Amounts.get(initialCurrency, 1.0)  # Start with the initial amount
    for currency in day1Amounts:
        amount = day1Amounts[currency]
        day2Amounts = getMaxAmounts(pairs2, rates2, currency)
        maxFinalAmount = max(maxFinalAmount, amount * day2Amounts.get(initialCurrency, 1.0))

    return maxFinalAmount
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