How do you solve Maximize Amount After Two Days of Conversions on LeetCode?

Maximize Amount After Two Days of Conversions (LeetCode #3387) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding currency conversion as a graph problem helps in visualizing the relationships between currencies.

What is the time complexity of Maximize Amount After Two Days of Conversions?

The optimal solution for Maximize Amount After Two Days of Conversions runs in O(n) time and uses O(n) space. The complexity is linear because we are only iterating through the pairs once to build the conversion maps.

What is the brute force solution for Maximize Amount After Two Days of Conversions?

The brute force approach for Maximize Amount After Two Days of Conversions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible conversion path for both days. This means we will explore all combinations of conversions from the initial currency to every possible intermediate currency and back. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Maximize Amount After Two Days of Conversions?

Think about how you can represent the problem as a graph or a mapping. Always check edge cases, such as no conversions or only one conversion available. Explain your thought process clearly as you work through the problem.

What are common mistakes when solving Maximize Amount After Two Days of Conversions?

Not considering all possible intermediate currencies. Failing to account for the conversion back to the initial currency.

#3387

Maximize Amount After Two Days of Conversions

Medium

ArrayStringDepth-First SearchBreadth-First SearchGraph TheoryHash MapGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a graph-like approach to find the best conversion rates. We can use a dictionary to store the maximum conversion rates for each currency, allowing us to efficiently compute the best conversions in linear time.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping of conversion rates for day 1 and day 2 using dictionaries.
2Step 2: For each currency in the day 1 mapping, calculate the maximum amount obtainable after day 2 conversions.
3Step 3: Return the maximum amount found.

solution.py17 lines

1def maxAmount(initialCurrency, pairs1, rates1, pairs2, rates2):
2    from collections import defaultdict
3    rates_day1 = defaultdict(float)
4    rates_day2 = defaultdict(float)
5    for i in range(len(pairs1)):
6        rates_day1[pairs1[i][0]] = max(rates_day1[pairs1[i][0]], rates1[i])
7    for i in range(len(pairs2)):
8        rates_day2[pairs2[i][0]] = max(rates_day2[pairs2[i][0]], rates2[i])
9    max_amount = 1.0
10    for currency in rates_day1:
11        if currency == initialCurrency:
12            continue
13        amount_day1 = 1.0 * rates_day1[currency]
14        if currency in rates_day2:
15            amount_day2 = amount_day1 * rates_day2[currency]
16            max_amount = max(max_amount, amount_day2)
17    return max_amount

ℹ

Complexity note: The complexity is linear because we are only iterating through the pairs once to build the conversion maps.

1Understanding currency conversion as a graph problem helps in visualizing the relationships between currencies.
2Using dictionaries to store maximum rates allows for efficient lookups and calculations.

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