Problem Description

You have n cities numbered from 1 to n. These cities are connected by bidirectional roads given in a 2D array roads, where roads[i] = [ai, bi, costi] means there's a road between city ai and city bi with travel cost costi.

Each city sells apples at different prices. The array appleCost tells you the price of an apple in each city, where appleCost[i] is the cost to buy one apple from city i (using 0-based indexing in the actual implementation).

Here's the task:

1. Start from any city
2. Travel through roads to reach any city (could be the starting city itself)
3. Buy exactly one apple from that city
4. Return back to your starting city

The catch is that after buying the apple, all road costs are multiplied by a factor k for your return journey. So if a road originally costs c, it will cost c * k on the way back.

Given the multiplier k, you need to find the minimum total cost for each possible starting city. Return an array answer where answer[i] represents the minimum cost to complete the entire journey (go somewhere, buy an apple, and return) if you start from city i.

The total cost for a journey includes:

• Cost to travel from starting city to the city where you buy the apple
• Cost of the apple itself
• Cost to travel back (with road costs multiplied by k)

Since the same roads are used for going and returning, if the shortest path from city A to city B costs d, then the total travel cost would be d + d * k = d * (1 + k).

Intuition

The key insight is recognizing that we need to find the optimal city to buy an apple from, for each possible starting city. Since we must return to our starting point after buying the apple, and the return journey costs are multiplied by k, the total travel cost becomes distance * (1 + k).

Think about what happens when we start from city i:

• We need to reach some city j to buy an apple
• The shortest path from i to j has some cost d
• We buy an apple at city j for appleCost[j]
• We return via the same shortest path, but now it costs d * k
• Total cost = d + appleCost[j] + d * k = d * (1 + k) + appleCost[j]

This means for each starting city, we want to minimize: distance_to_city * (1 + k) + apple_cost_at_city

Since we need to evaluate this for every possible destination city from each starting point, we need the shortest distances from each starting city to all other cities. This naturally leads us to use Dijkstra's algorithm.

Why Dijkstra? Because:

1. We need shortest paths from a single source (the starting city) to all other cities
2. All edge weights (road costs) are positive
3. We can calculate the minimum cost on-the-fly as we explore cities

During the Dijkstra traversal from starting city i, when we reach any city u with shortest distance d, we can immediately calculate the total cost if we buy an apple there: d * (k + 1) + appleCost[u]. We keep track of the minimum among all such costs.

The beauty of this approach is that Dijkstra guarantees we visit each city with its shortest distance from the source, so we can confidently calculate the cost for buying an apple at each city and maintain the global minimum.

Learn more about Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The solution implements a heap-optimized Dijkstra's algorithm for each city as the starting point. Here's how the implementation works:

Graph Construction: First, we build an adjacency list representation of the graph:

g = defaultdict(list)
for a, b, c in roads:
    a, b = a - 1, b - 1  # Convert to 0-based indexing
    g[a].append((b, c))
    g[b].append((a, c))

Since roads are bidirectional, we add edges in both directions. The cities are converted from 1-based to 0-based indexing for easier array manipulation.

Dijkstra's Algorithm for Each Starting City: For each city i from 0 to n-1, we run a modified Dijkstra's algorithm:

def dijkstra(i):
    q = [(0, i)]  # Priority queue: (distance, city)
    dist = [inf] * n  # Shortest distances from city i
    dist[i] = 0
    ans = inf  # Minimum cost to complete the journey

The algorithm uses:

• A min-heap q to always process the city with minimum distance first
• A dist array to track the shortest distance from starting city i to all other cities
• An ans variable to track the minimum total cost found so far

Core Algorithm Logic:

while q:
    d, u = heappop(q)
    ans = min(ans, appleCost[u] + d * (k + 1))
    for v, w in g[u]:
        if dist[v] > dist[u] + w:
            dist[v] = dist[u] + w
            heappush(q, (dist[v], v))

For each city u we visit:

1. Calculate the total cost if we buy an apple here: appleCost[u] + d * (k + 1)
   • d is the shortest distance from starting city to u
   • d * (k + 1) represents the round-trip cost (going costs d, returning costs d * k)
2. Update ans with the minimum cost found
3. Explore neighbors and update their distances if we found a shorter path
4. Add updated neighbors to the priority queue

Final Result:

return [dijkstra(i) for i in range(n)]

We run Dijkstra from each city and collect all minimum costs in a list.

Time Complexity: O(n * (n + m) * log n) where n is the number of cities and m is the number of roads. We run Dijkstra n times, and each run takes O((n + m) * log n) with the heap operations.

Space Complexity: O(n + m) for storing the graph and the distance array.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• n = 4 cities (numbered 1 to 4)
• roads = [[1,2,4], [2,3,2], [2,4,5], [3,4,1]]
• appleCost = [56, 42, 86, 23] (0-indexed: city 1 costs 56, city 2 costs 42, etc.)
• k = 2 (return journey costs are doubled)

Graph Visualization:

    1 --4-- 2
           /|\
          2 | 5
           \|/
            3 --1-- 4

Step 1: Convert to 0-based indexing and build adjacency list

City 0: [(1, 4)]
City 1: [(0, 4), (2, 2), (3, 5)]
City 2: [(1, 2), (3, 1)]
City 3: [(1, 5), (2, 1)]

Step 2: Run Dijkstra from City 0 (originally City 1)

Initialize:

• Priority queue: [(0, 0)] (distance 0 to itself)
• Distances: [0, inf, inf, inf]
• Minimum cost: inf

Processing:

1. Pop (0, 0) - at city 0 with distance 0

   • Cost if buying here: 56 + 0*(2+1) = 56
   • Update minimum: ans = 56
   • Explore neighbor city 1: update distance to 4, push (4, 1)

2. Pop (4, 1) - at city 1 with distance 4

   • Cost if buying here: 42 + 4*3 = 54
   • Update minimum: ans = 54
   • Explore neighbors:
     • City 2: distance becomes 6, push (6, 2)
     • City 3: distance becomes 9, push (9, 3)

3. Pop (6, 2) - at city 2 with distance 6

   • Cost if buying here: 86 + 6*3 = 104
   • Minimum stays: ans = 54
   • Explore city 3: distance becomes 7 (better than 9), push (7, 3)

4. Pop (7, 3) - at city 3 with distance 7

   • Cost if buying here: 23 + 7*3 = 44
   • Update minimum: ans = 44

Result for starting from city 0: 44 (buy apple at city 3)

Step 3: Repeat for other starting cities

Starting from City 1 (originally City 2):

• Shortest paths: [4, 0, 2, 3]
• Costs: City 0: 56+12=68, City 1: 42+0=42, City 2: 86+6=92, City 3: 23+9=32
• Minimum: 32

Starting from City 2 (originally City 3):

• Shortest paths: [6, 2, 0, 1]
• Costs: City 0: 56+18=74, City 1: 42+6=48, City 2: 86+0=86, City 3: 23+3=26
• Minimum: 26

Starting from City 3 (originally City 4):

• Shortest paths: [7, 3, 1, 0]
• Costs: City 0: 56+21=77, City 1: 42+9=51, City 2: 86+3=89, City 3: 23+0=23
• Minimum: 23

Final Answer: [44, 32, 26, 23]

This means:

• Starting from city 1: minimum cost is 44 (go to city 4, buy apple for 23, return)
• Starting from city 2: minimum cost is 32 (go to city 4, buy apple for 23, return)
• Starting from city 3: minimum cost is 26 (go to city 4, buy apple for 23, return)
• Starting from city 4: minimum cost is 23 (buy apple locally, no travel needed)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × m × log m)

The algorithm runs Dijkstra's algorithm from each of the n cities. For each Dijkstra execution:

• The while loop processes each edge at most once, giving us O(m) edge relaxations
• Each edge relaxation involves a heap push operation, which costs O(log |heap|)
• The heap can contain at most m elements (since we can push at most once per edge relaxation)
• Therefore, each Dijkstra run takes O(m × log m) time

Since we run Dijkstra n times (once for each starting city), the total time complexity is O(n × m × log m).

Space Complexity: O(n + m)

The space usage consists of:

• Graph adjacency list g: stores all edges, requiring O(m) space (each edge is stored twice for undirected graph)
• Distance array dist: requires O(n) space for each Dijkstra run
• Priority queue q: can contain at most O(m) elements in the worst case
• Result list: requires O(n) space

Since these data structures don't scale multiplicatively and the Dijkstra function reuses the same space for each iteration, the overall space complexity is O(n + m).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Incorrect Distance Check in Dijkstra's Algorithm

The Problem: The current implementation has a subtle but critical bug in the Dijkstra's algorithm. When we pop a node from the priority queue, we don't verify if we've already processed this node with a shorter distance. This can lead to processing the same node multiple times with outdated distances, causing incorrect results.

# Current problematic code:
while priority_queue:
    current_distance, current_city = heappop(priority_queue)
    # No check if this is an outdated entry!
    total_cost = appleCost[current_city] + current_distance * (k + 1)
    min_apple_cost = min(min_apple_cost, total_cost)

Why This Happens: In Dijkstra's algorithm with a min-heap, we might push the same city multiple times with different distances. When we later pop an outdated entry (with a larger distance), we shouldn't process it again.

The Solution: Add a check to skip processing if the current distance is greater than the already recorded shortest distance:

while priority_queue:
    current_distance, current_city = heappop(priority_queue)
  
    # Skip if this is an outdated entry
    if current_distance > shortest_distances[current_city]:
        continue
  
    total_cost = appleCost[current_city] + current_distance * (k + 1)
    min_apple_cost = min(min_apple_cost, total_cost)

Pitfall 2: Integer Overflow with Large k Values

The Problem: When calculating current_distance * (k + 1), if both current_distance and k are large, the multiplication might overflow in languages with fixed integer sizes. While Python handles arbitrary precision integers, this is a common issue when porting to other languages.

The Solution: Be mindful of the constraints and consider using appropriate data types or adding overflow checks:

# For very large values, consider checking bounds
total_cost = appleCost[current_city] + current_distance * (k + 1)
if total_cost > 10**15:  # Or some reasonable upper bound
    continue  # This path is too expensive anyway

Pitfall 3: Not Handling Edge Cases

The Problem: The code doesn't explicitly handle certain edge cases:

• What if a city has no roads connecting to it? (isolated city)
• What if k = 0? (no extra cost for return journey)

The Solution: The current implementation actually handles these cases correctly:

• Isolated cities will only consider buying an apple from themselves (distance = 0)
• When k = 0, the formula d * (k + 1) = d * 1 = d correctly represents just the one-way cost

However, it's good practice to document or test these edge cases explicitly:

# Edge case: buying apple from starting city itself
# This is automatically handled as shortest_distances[start_city] = 0
# Cost = appleCost[start_city] + 0 * (k + 1) = appleCost[start_city]

Complete Corrected Solution:

def dijkstra(start_city: int) -> int:
    priority_queue = [(0, start_city)]
    shortest_distances = [inf] * n
    shortest_distances[start_city] = 0
    min_apple_cost = inf
  
    while priority_queue:
        current_distance, current_city = heappop(priority_queue)
      
        # Critical fix: Skip outdated entries
        if current_distance > shortest_distances[current_city]:
            continue
      
        total_cost = appleCost[current_city] + current_distance * (k + 1)
        min_apple_cost = min(min_apple_cost, total_cost)
      
        for neighbor_city, road_cost in adjacency_list[current_city]:
            new_distance = shortest_distances[current_city] + road_cost
          
            if shortest_distances[neighbor_city] > new_distance:
                shortest_distances[neighbor_city] = new_distance
                heappush(priority_queue, (new_distance, neighbor_city))
  
    return min_apple_cost