Minimum Cost to Buy Apples

class Solution {
public:
    vector<long long> calculateMinCost(int count, vector<vector<int>>& routes, vector<int>& fruitCost, int multiplier) {
        vector<vector<pair<int, int>>> adjacencyList(count);
    
        for (auto& route : routes) {
            int from = route[0] - 1, to = route[1] - 1, travelCost = route[2];
            adjacencyList[from].push_back({to, travelCost});
            adjacencyList[to].push_back({from, travelCost});
        }
    
        vector<long long> minCosts(fruitCost.begin(), fruitCost.end());
    
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> minHeap;
        for (int i = 0; i < fruitCost.size(); i++) {
            minHeap.push({fruitCost[i], i});
        }
    
        while (!minHeap.empty()) {
            auto [currentCost, currentCity] = minHeap.top();
            minHeap.pop();
    
            if (minCosts[currentCity] < currentCost)
                continue;
    
            for (auto& neighbor : adjacencyList[currentCity]) {
                int nextCity = neighbor.first;
                int nextCost = neighbor.second;
    
                if (minCosts[nextCity] > minCosts[currentCity] + (multiplier + 1) * nextCost) {
                    minCosts[nextCity] = minCosts[currentCity] + (multiplier + 1) * nextCost;
                    minHeap.push({minCosts[nextCity], nextCity});
                }
            }
        }
    
        return minCosts;
    }
};

This solution describes the approach to minimize the cost of buying apples across multiple cities using a specific strategy that involves city connections, travel costs, and fruit prices.

Here's a concise overview of how the provided C++ code executes:

• Begin by constructing an adjacency list from the provided routes between cities. Each entry contains pairs indicating the neighboring city and the cost to travel there.

• Initialize a vector, minCosts, based on initial fruit costs in each city. This vector is pivotal in tracking the minimal cost to purchase and transport apples from any city.

• Employ a min-heap (priority queue) to facilitate an efficient way to explore cities with the currently lowest cost.

• Use a Dijkstra-esque algorithm modified with a given cost multiplier:

  1. Extract the city with the minimum cost.
  2. Ignore if the new cost fetched from the heap for this city is higher than the current known minimum (this indicates another route found a cheaper way).
  3. Update the cost for neighboring cities if traveling from the current city results in a cheaper total cost. This update integrates both travel and apple cost tailored by the multiplier.

• The logic incorporates optimization to prevent unnecessary computations, ensuring that only promising routes (those potentially lowering the total costs) are pursued further.

This implementation effectively combines graph traversal and cost analysis, optimizing the total expenditure required to purchase and transport apples while taking different costs and routes into consideration. The use of data structures like vectors, pairs, and priority queues is crucial in managing the complexity and improving the efficiency of the solution.

class Solution {
    public long[] calculateMinFruitCost(int cityCount, int[][] connections, int[] costs, int multiplier) {
        List<List<Pair<Integer, Integer>>> cityGraph = new ArrayList<>();
        for (int i = 0; i < cityCount; ++i) {
            cityGraph.add(new ArrayList<>());
        }
            
        for (int[] connection : connections) {
            int first = connection[0] - 1, second = connection[1] - 1, travelCost = connection[2];
            cityGraph.get(first).add(new Pair<Integer, Integer>(second, travelCost));
            cityGraph.get(second).add(new Pair<Integer, Integer>(first, travelCost));
        }
    
        long[] minCosts = new long[cityCount];
        for (int i = 0; i < cityCount; i++) {
            minCosts[i] = costs[i];
        }
            
        Queue<Pair<Long, Integer>> priorityQueue = new PriorityQueue<>((x, y) -> 
                Long.compare(x.getKey(), y.getKey()));
        for (int start = 0; start < cityCount; start++) {
            priorityQueue.add(new Pair<>((long)costs[start], start));
        }
            
        while (!priorityQueue.isEmpty()) {
            Pair<Long, Integer> currentPair = priorityQueue.poll();
            long currentCost = currentPair.getKey();
            int currentCity = currentPair.getValue();
                
            if (minCosts[currentCity] < currentCost)
                continue;
                
            for (Pair<Integer, Integer> edge : cityGraph.get(currentCity)) {
                int neighborCity = edge.getKey(), edgeCost = edge.getValue();
                if (minCosts[neighborCity] > minCosts[currentCity] + (multiplier + 1) * edgeCost) {
                    minCosts[neighborCity] = minCosts[currentCity] + (multiplier + 1) * edgeCost;
                    priorityQueue.add(new Pair<Long, Integer>(minCosts[neighborCity], neighborCity));
                }
            }
        }
        return minCosts;
    }
}

The provided Java solution determines the minimal costs for purchasing apples across multiple cities that are connected by specified routes. Here is a step-by-step breakdown of the solution's strategy:

1. Initialize a graph where each city is represented as a node and each connection between cities (with associated travel costs) forms an edge in the graph.

2. Use a priority queue to efficiently process cities starting from the one with the lowest apple buying cost.

3. Calculate the minimum cost to buy apples for each city by considering both direct apple costs at a city and the transportation costs to travel from already processed cities with potentially cheaper apples.

4. Leverage a modification of Dijkstra's algorithm to populate the results:

   • While the priority queue is not empty, dequeue the city with the current lowest known cost.
   • Update the costs to reach neighboring cities through this city, taking into account the travel cost and a specified multiplier that adjusts this cost.
   • If a cheaper route to a neighboring city is found, update the cost and enqueue this city in the priority queue for further processing.

This approach ensures that the resulting array, minCosts, contains the least cost required for purchasing apples from the most optimal city including travel costs. This operation is completed in an efficient manner given the use of a priority queue, which enables more immediate processing of the potentially cheaper routes.

class RouteFinder:
    def findCheapest(self, city_count: int, pathways: List[List[int]], buyCost: List[int], multiplier: int) -> List[int]:
        city_graph = [[] for _ in range(city_count)]
            
        for start, end, expense in pathways:
            city_graph[start - 1].append((end - 1, expense))
            city_graph[end - 1].append((start - 1, expense))
            
        cheapest = list(buyCost)
            
        priority = [(cost, city) for city, cost in enumerate(buyCost)]
        heapq.heapify(priority)
            
        while priority:
            accumulated_cost, current_city = heapq.heappop(priority)
                
            if cheapest[current_city] < accumulated_cost:
                continue
                
            for adjacent, transit_cost in city_graph[current_city]:
                new_cost = cheapest[current_city] + (multiplier + 1) * transit_cost
                if cheapest[adjacent] > new_cost:
                    cheapest[adjacent] = new_cost
                    heapq.heappush(priority, (cheapest[adjacent], adjacent))
            
        return cheapest

The provided Python code defines a class RouteFinder that offers a method to determine the minimum cost of purchasing apples across multiple cities. This method, findCheapest, takes as arguments the number of cities, a list of pathways (each represented by a trio of values: start city, end city, and pathway expense), a list indicating the cost to buy apples directly in each city, and a multiplier affecting the cost of traveling between cities.

Understand the process step-by-step:

1. Initialize a graph where each city points to its directly connected neighbors and the associated travel costs.
2. Use Dijkstra's algorithm strategy with a priority queue (min-heap) to explore the lowest cost pathways first.
3. Begin with the direct buyCost for each city as the initial estimated cheapest cost.
4. Iteratively, calculate a new potential cost for each connected city factoring in travel costs modified by the multiplier.
5. Whenever a cheaper cost is found, update the priority queue and the cheapest cost array.
6. The result is an array where each index represents a city and the value at each index is the minimum cost to buy apples in that city considering both direct purchases and costs incurred by traveling through other cities first.

This method uses efficient data structures and algorithms to ensure that we can find the minimum cost of buying apples in a scenario where cities are interconnected and costs can vary based on travel paths.