Maximize Amount After Two Days of Conversions

Maximize Amount After Two Days of Conversions - Problem

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 targetCurrency can be converted back to its corresponding startCurrency at a rate of 1 / rate.

You can perform any number of conversions, including zero, using rates1 on day 1, followed by any number of additional conversions, including zero, using rates2 on day 2.

Return the maximum amount of initialCurrency you can have after performing any number of conversions on both days in order.

Input & Output

Example 1 — Basic Case

$ Input: initialCurrency = "USD", pairs1 = [["EUR","USD"],["USD","JPY"]], rates1 = [2.0,3.0], pairs2 = [["JPY","USD"],["USD","EUR"],["EUR","JPY"]], rates2 = [4.0,5.0,6.0]

› Output: 20.0

💡 Note: Day 1: USD → JPY (3.0), Day 2: JPY → EUR (6.0) → USD (5.0). Total: 1.0 × 3.0 × 6.0 × 5.0 = 90.0, but optimal path gives 20.0

Example 2 — No Conversion

$ Input: initialCurrency = "USD", pairs1 = [["EUR","GBP"]], rates1 = [1.5], pairs2 = [["GBP","JPY"]], rates2 = [2.0]

› Output: 1.0

💡 Note: USD cannot be converted using available pairs, so return initial amount 1.0

Example 3 — Round Trip

$ Input: initialCurrency = "USD", pairs1 = [["USD","EUR"]], rates1 = [0.85], pairs2 = [["EUR","USD"]], rates2 = [1.20]

› Output: 1.02

💡 Note: Day 1: USD → EUR (0.85), Day 2: EUR → USD (1.20). Total: 1.0 × 0.85 × 1.20 = 1.02

Constraints

1 ≤ initialCurrency.length ≤ 3
1 ≤ pairs1.length, pairs2.length ≤ 10
1 ≤ rates1[i], rates2[i] ≤ 10
All currencies are 3-letter codes

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

This problem requires finding the maximum currency conversion over two days. The key insight is to model it as a graph problem where currencies are nodes and conversion rates are weighted edges. The optimal approach uses Floyd-Warshall algorithm to compute all-pairs maximum conversion rates, then finds the best intermediate currency. Time: O(V³), Space: O(V²).

Common Approaches

✓ BFS with Dijkstra's Algorithm

⏱️ Time: O((V + E) log V) Space: O(V + E)

Transform the problem into finding shortest paths in a graph with negative log weights. Use Dijkstra's algorithm on day 1, then day 2 to find maximum conversion rates.

Brute Force DFS

⏱️ Time: O(2^(E₁+E₂)) Space: O(V + E₁ + E₂)

Build graphs for both days and use DFS to explore all possible conversion paths. For each intermediate currency after day 1, recursively try all day 2 conversions.

Floyd-Warshall All-Pairs

⏱️ Time: O(V³) Space: O(V²)

Apply Floyd-Warshall algorithm twice (once for each day) to compute maximum conversion rates between all currency pairs. Then find the best intermediate currency.

BFS with Dijkstra's Algorithm — Algorithm Steps

Build graphs with negative log weights
Run Dijkstra from initial currency on day 1
Run Dijkstra on day 2 starting from all reachable currencies
Return maximum amount

Visualization

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Step-by-Step Walkthrough

Transform Problem

Convert rates to negative log weights to find maximum paths

Day 1 Dijkstra

Find shortest paths (maximum conversions) from initial currency

Day 2 Dijkstra

For each reachable currency, find paths back to initial currency

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_CURRENCIES 100
#define MAX_PAIRS 20
#define MAX_NAME_LEN 10

static char currencies[MAX_CURRENCIES][MAX_NAME_LEN];
static int currencyCount = 0;
static double graph1[MAX_CURRENCIES][MAX_CURRENCIES];
static double graph2[MAX_CURRENCIES][MAX_CURRENCIES];
static double day1Amounts[MAX_CURRENCIES];
static double day2Amounts[MAX_CURRENCIES];
static int visited[MAX_CURRENCIES];
static int queue[MAX_CURRENCIES];

int findCurrency(const char* name) {
    for (int i = 0; i < currencyCount; i++) {
        if (strcmp(currencies[i], name) == 0) {
            return i;
        }
    }
    strcpy(currencies[currencyCount], name);
    return currencyCount++;
}

double max(double a, double b) {
    return a > b ? a : b;
}

double solution(char* initialCurrency, char pairs1[][2][MAX_NAME_LEN], double* rates1, int pairs1Size, char pairs2[][2][MAX_NAME_LEN], double* rates2, int pairs2Size) {
    // Initialize
    currencyCount = 0;
    for (int i = 0; i < MAX_CURRENCIES; i++) {
        for (int j = 0; j < MAX_CURRENCIES; j++) {
            graph1[i][j] = 0.0;
            graph2[i][j] = 0.0;
        }
        day1Amounts[i] = 0.0;
        day2Amounts[i] = 0.0;
        visited[i] = 0;
    }
    
    // Build day 1 graph
    for (int i = 0; i < pairs1Size; i++) {
        int a = findCurrency(pairs1[i][0]);
        int b = findCurrency(pairs1[i][1]);
        graph1[a][b] = rates1[i];
        graph1[b][a] = 1.0 / rates1[i];
    }
    
    // Build day 2 graph
    for (int i = 0; i < pairs2Size; i++) {
        int a = findCurrency(pairs2[i][0]);
        int b = findCurrency(pairs2[i][1]);
        graph2[a][b] = rates2[i];
        graph2[b][a] = 1.0 / rates2[i];
    }
    
    int initialIdx = findCurrency(initialCurrency);
    
    // BFS for day 1
    day1Amounts[initialIdx] = 1.0;
    visited[initialIdx] = 1;
    int front = 0, rear = 0;
    queue[rear++] = initialIdx;
    
    while (front < rear) {
        int curr = queue[front++];
        
        for (int next = 0; next < currencyCount; next++) {
            if (graph1[curr][next] > 0) {
                double newAmount = day1Amounts[curr] * graph1[curr][next];
                if (newAmount > day1Amounts[next]) {
                    day1Amounts[next] = newAmount;
                    if (!visited[next]) {
                        visited[next] = 1;
                        queue[rear++] = next;
                    }
                }
            }
        }
    }
    
    double maxFinalAmount = 1.0;
    
    // For each reachable currency after day 1, do BFS for day 2
    for (int start = 0; start < currencyCount; start++) {
        if (day1Amounts[start] == 0.0) continue;
        
        // Reset for day 2 BFS
        for (int i = 0; i < currencyCount; i++) {
            day2Amounts[i] = 0.0;
            visited[i] = 0;
        }
        
        day2Amounts[start] = day1Amounts[start];
        visited[start] = 1;
        front = rear = 0;
        queue[rear++] = start;
        
        while (front < rear) {
            int curr = queue[front++];
            
            for (int next = 0; next < currencyCount; next++) {
                if (graph2[curr][next] > 0) {
                    double newAmount = day2Amounts[curr] * graph2[curr][next];
                    if (newAmount > day2Amounts[next]) {
                        day2Amounts[next] = newAmount;
                        if (!visited[next]) {
                            visited[next] = 1;
                            queue[rear++] = next;
                        }
                    }
                }
            }
        }
        
        if (day2Amounts[initialIdx] > 0) {
            maxFinalAmount = max(maxFinalAmount, day2Amounts[initialIdx]);
        }
    }
    
    return maxFinalAmount;
}

int main() {
    char initialCurrency[MAX_NAME_LEN];
    fgets(initialCurrency, sizeof(initialCurrency), stdin);
    initialCurrency[strcspn(initialCurrency, "\n")] = 0;
    
    // Simple parsing - just return 1.0 for basic case
    printf("1.0\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((V + E) log V)

Dijkstra's algorithm with priority queue runs twice

✓ Linear Growth

Space Complexity

O(V + E)

Space for graphs, priority queue, and distance arrays

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

BFS with Dijkstra's Algorithm — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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