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

The optimal solution has a time complexity of O(E1 + E2 + V).

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

The space complexity is O(V + E1 + E2).

What data structures are used to solve Maximize Amount After Two Days of Conversions?

This problem uses Array, String, Depth-First Search, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximize Amount After Two Days of Conversions?

Insight 1: The problem can be modeled as finding the best path in a graph where nodes represent currencies and edges represent conversion rates. The order of conversions matters (day 1 then day 2). Insight 2: We need to find the maximum rate achievable from the initial currency to any other currency on day 1, and then from that currency back to the initial currency on day 2 (in reverse).

Which companies ask LeetCode 3387. Maximize Amount After Two Days of Conversions in interviews?

This problem is frequently asked at Rippling.

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Maximize Amount After Two Days of Conversions - LeetCode 3387 Solution

Maximize Amount After Two Days of Conversions - Complete Solution Guide

Maximize Amount After Two Days of Conversions is LeetCode problem 3387, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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[i] = [startCurrency i , targetCurrency i ] denotes that you can convert from startCurrency i to targetCurrency i at a rate of rates1[i] on day 1 . pairs2[i] = [startCurrency i , targetCurrency i ] denotes that you can convert from startCurrency i to targetCurrency i at a rate of rates2[i] on day 2 . Also, each targetCur

Detailed Explanation

The problem requires you to find the maximum amount of a given `initialCurrency` you can obtain after two days of currency conversions. You start with 1.0 of the `initialCurrency`. On day 1, you can convert currencies using `pairs1` and `rates1`. On day 2, you use `pairs2` and `rates2`. You can convert from currency A to currency B at a specific rate and also back from B to A at the inverse rate (1/rate). The goal is to strategically convert currencies over these two days to maximize the final amount in terms of the initial currency.

Solution Approach

The provided solution uses Breadth-First Search (BFS) to find the maximum conversion rates. It first constructs adjacency lists representing the possible conversions for each day. BFS is then performed twice: once for day 1 to calculate the maximum conversion rates from the `initialCurrency` to all other currencies and once (in reverse) for day 2 to calculate the maximum rate of conversion from all currencies back to initialCurrency.

Step-by-Step Algorithm

Step 1: Build Adjacency Lists: For each day, construct an adjacency list (or graph) where nodes represent currencies, and edges represent conversion rates between currencies. Add both forward and reverse conversion edges (using the inverse rate).
Step 2: Run BFS for Day 1: Perform BFS starting from the `initialCurrency`. During the traversal, keep track of the maximum conversion rate from the `initialCurrency` to each reachable currency. Store these rates in a `rates1_from_initial` map.
Step 3: Run BFS for Day 2 (Reversed Conversion): Since the amount you get at end of Day 1 is the starting amount for Day 2, and you need the conversions from Day 2 to eventually end up at Day 1's initial currency to compare overall rates, perform BFS from each Day 1 reachable currency. Since each currency from Day 1 can be reached at a certain rate (`rates1_from_initial`), we consider those final day 1 currencies as starting currencies for Day 2. In other words, we perform BFS for Day 2 to get `rates2_from_initial` for each currency in day 2.
Step 4: Calculate the Maximum Amount: Iterate through all the currencies encountered in both days. For each currency, calculate the product of the conversion rate from `initialCurrency` on day 1 to that currency and the reverse conversion rate on day 2 from that currency back to initial currency. The maximum amount achievable is the maximum of these products.
Step 5: Return the Maximum Amount: Return the calculated maximum amount.

Key Insights

Insight 1: The problem can be modeled as finding the best path in a graph where nodes represent currencies and edges represent conversion rates. The order of conversions matters (day 1 then day 2).
Insight 2: We need to find the maximum rate achievable from the initial currency to any other currency on day 1, and then from that currency back to the initial currency on day 2 (in reverse).
Insight 3: Breadth-First Search (BFS) is suitable for exploring all possible conversion paths and determining the maximum conversion rate achievable from the initial currency to all other currencies in each day's conversions.

Complexity Analysis

Time Complexity: O(E1 + E2 + V)

Space Complexity: O(V + E1 + E2)

Topics

This problem involves: Array, String, Depth-First Search, Breadth-First Search, Graph.

Companies

Asked at: Rippling.

Maximize Amount After Two Days of Conversions | LeetCode 3387

Medium

Array String Depth-First Search Breadth-First Search Graph

Time: O(E1 + E2 + V)Space: O(V + E1 + E2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public double maxAmount(String initialCurrency, List<List<String>> pairs1, double[] rates1, List<List<String>> pairs2, double[] rates2) {
        
        class Helper {
            public Map<String, Double> run_bfs(String start_node, List<List<String>> pairs, double[] rates) {
                Map<String, List<Pair<String, Double>>> adj = new HashMap<>();
                for (int i = 0; i < pairs.size(); ++i) {
                    List<String> edge = pairs.get(i);
                    String u = edge.get(0);
                    String v = edge.get(1);
                    double rate = rates[i];

                    adj.computeIfAbsent(u, k -> new ArrayList<>()).add(new Pair<>(v, rate));
                    adj.computeIfAbsent(v, k -> new ArrayList<>()).add(new Pair<>(u, 1.0 / rate));
                }

                Map<String, Double> rates_from_start = new HashMap<>();
                rates_from_start.put(start_node, 1.0);
                Queue<Pair<String, Double>> q = new LinkedList<>();
                q.offer(new Pair<>(start_node, 1.0));

                while (!q.isEmpty()) {
                    Pair<String, Double> curr_pair = q.poll();
                    String curr = curr_pair.getKey();
                    double rate_from_start = curr_pair.getValue();

                    if (adj.containsKey(curr)) {
                        for (Pair<String, Double> neighbor_pair : adj.get(curr)) {
                            String neighbor = neighbor_pair.getKey();
                            double edge_rate = neighbor_pair.getValue();

                            if (!rates_from_start.containsKey(neighbor)) {
                                double new_rate = rate_from_start * edge_rate;
                                rates_from_start.put(neighbor, new_rate);
                                q.offer(new Pair<>(neighbor, new_rate));
                            }
                        }
                    }
                }

                return rates_from_start;
            }
        }

        Helper helper = new Helper();

        Set<String> all_currencies = new HashSet<>();
        all_currencies.add(initialCurrency);
        for (List<String> edge : pairs1) {
            all_currencies.add(edge.get(0));
            all_currencies.add(edge.get(1));
        }
        for (List<String> edge : pairs2) {
            all_currencies.add(edge.get(0));
            all_currencies.add(edge.get(1));
        }

        Map<String, Double> rates1_from_initial = helper.run_bfs(initialCurrency, pairs1, rates1);
        Map<String, Double> rates2_from_initial = helper.run_bfs(initialCurrency, pairs2, rates2);

        double max_amount = 0.0;

        for (String c_intermediate : all_currencies) {
            double rate1 = rates1_from_initial.getOrDefault(c_intermediate, 0.0);

            Double rate2_from_init_obj = rates2_from_initial.get(c_intermediate);
            double rate2_from_init = rate2_from_init_obj == null ? 0.0 : rate2_from_init_obj;
            double rate2 = (rate2_from_init > 0.0) ? (1.0 / rate2_from_init) : 0.0;

            double current_amount = rate1 * rate2;
            max_amount = Math.max(max_amount, current_amount);
        }

        return max_amount;
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 3387: Maximize Amount After Two Days of Conversions - Detailed explanation and solution approaches.

Topics

Array String Depth-First Search Breadth-First Search Graph

Companies

Rippling

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3387

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Build Adjacency Lists: For each day, construct an adjacency list (or graph) where nodes represent currencies, and edges represent conversion rates between currencies. Add both forward and reverse conversion edges (using the inverse rate).
2
Step 2: Run BFS for Day 1: Perform BFS starting from the `initialCurrency`. During the traversal, keep track of the maximum conversion rate from the `initialCurrency` to each reachable currency. Store these rates in a `rates1_from_initial` map.
3
Step 3: Run BFS for Day 2 (Reversed Conversion): Since the amount you get at end of Day 1 is the starting amount for Day 2, and you need the conversions from Day 2 to eventually end up at Day 1's initial currency to compare overall rates, perform BFS from each Day 1 reachable currency. Since each currency from Day 1 can be reached at a certain rate (`rates1_from_initial`), we consider those final day 1 currencies as starting currencies for Day 2. In other words, we perform BFS for Day 2 to get `rates2_from_initial` for each currency in day 2.
4
Step 4: Calculate the Maximum Amount: Iterate through all the currencies encountered in both days. For each currency, calculate the product of the conversion rate from `initialCurrency` on day 1 to that currency and the reverse conversion rate on day 2 from that currency back to initial currency. The maximum amount achievable is the maximum of these products.
5
Step 5: Return the Maximum Amount: Return the calculated maximum amount.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(E1 + E2 + V)

💾

Space Complexity

O(V + E1 + E2)

📝

Why?

The time complexity is driven by the BFS traversals of the currency conversion graphs. Each BFS visits all currencies (V) and explores all conversion pairs (E). The space complexity is due to storing the adjacency lists (graphs) and the intermediate conversion rates.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bellman-Ford Algorithm

Instead of BFS, the Bellman-Ford algorithm could be used to find the maximum conversion rates. This is generally used for detecting negative cycles, but can be adapted.

Time: O(V*E), where V is the number of currencies and E is the total number of currency pairs across both days.Space: O(V), to store the distances (conversion rates) from the initial currency.

Trade-offs: Bellman-Ford is generally slower than BFS (especially since there are no negative cycles as per constraints), so BFS is more efficient. BFS's advantage is further amplified by the fact that the conversion graph is unweighted (all rates are positive).

Floyd-Warshall Algorithm

Floyd-Warshall algorithm can be used to find the shortest path between all pairs of currencies. Apply it for each day. Then iterate over intermediate nodes and determine the max amount.

Time: O(V^3), where V is number of currenciesSpace: O(V^2)

Trade-offs: Floyd-Warshall is slower than BFS. Also it takes up more memory. Not well suited for this problem.

✨

Key Insights

1Insight 1: The problem can be modeled as finding the best path in a graph where nodes represent currencies and edges represent conversion rates. The order of conversions matters (day 1 then day 2).
2Insight 2: We need to find the maximum rate achievable from the initial currency to any other currency on day 1, and then from that currency back to the initial currency on day 2 (in reverse).
3Insight 3: Breadth-First Search (BFS) is suitable for exploring all possible conversion paths and determining the maximum conversion rate achievable from the initial currency to all other currencies in each day's conversions.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: No conversions possible (no pairs). In this case, the BFS will only visit the initial currency, and the other rates will be 0. The algorithm correctly handles this, resulting in a final amount of 1.0.
Edge case 2: No path exists from the initial currency to some other currency. The BFS won't reach those currencies, effectively treating their rates as 0, and those currencies won't impact the final maximum amount calculation.
Edge case 3: Cases where inverse conversion is more profitable than direct conversions. The bidirectional nature of the graph (forward and reverse edges) in BFS ensures that the algorithm considers all profitable conversion paths.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the inverse conversion rates. Remember that you can always convert back to the original currency, and sometimes that conversion is more profitable than the forward conversion.

✗Common mistake 2: Not realizing that conversions happen sequentially (day 1 then day 2). You need to find the best currency to convert to on day 1, and then convert back to the initial currency (or another currency) on day 2. The total conversion rate is the product of the two rates, not their sum.

✗Common mistake 3: Confusing the pairs and rates arrays, or incorrectly constructing the adjacency list. Ensure correct graph construction with the proper rates

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Currency arbitrage. The algorithm can be used to identify opportunities to profit from discrepancies in currency exchange rates across different markets.

🏦Real-world scenario 2: Finding the best flight routes based on reward points conversions. Different airlines and credit card companies offer varying conversion rates for reward points. This algorithm could be used to determine the best way to convert points to maximize flight value.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximize Amount After Two Days of Conversions"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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