Leetcode #2473: Minimum Cost to Buy Apples
In this guide, we solve Leetcode #2473 Minimum Cost to Buy Apples in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads, where roads[i] = [ai, bi, costi] indicates that there is a bidirectional road between cities ai and bi with a cost of traveling equal to costi.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Graph, Array, Shortest Path, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: n = 4, roads = [[1,2,4],[2,3,2],[2,4,5],[3,4,1],[1,3,4]], appleCost = [56,42,102,301], k = 2
Output: [54,42,48,51]
Explanation: The minimum cost for each starting city is the following:
- Starting at city 1: You take the path 1 -> 2, buy an apple at city 2, and finally take the path 2 -> 1. The total cost is 4 + 42 + 4 * 2 = 54.
- Starting at city 2: You directly buy an apple at city 2. The total cost is 42.
- Starting at city 3: You take the path 3 -> 2, buy an apple at city 2, and finally take the path 2 -> 3. The total cost is 2 + 42 + 2 * 2 = 48.
- Starting at city 4: You take the path 4 -> 3 -> 2 then you buy at city 2, and finally take the path 2 -> 3 -> 4. The total cost is 1 + 2 + 42 + 1 * 2 + 2 * 2 = 51.
Python Solution
class Solution:
def minCost(
self, n: int, roads: List[List[int]], appleCost: List[int], k: int
) -> List[int]:
def dijkstra(i):
q = [(0, i)]
dist = [inf] * n
dist[i] = 0
ans = inf
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))
return ans
g = defaultdict(list)
for a, b, c in roads:
a, b = a - 1, b - 1
g[a].append((b, c))
g[b].append((a, c))
return [dijkstra(i) for i in range(n)]
Complexity
The time complexity is , where and are the number of cities and roads, respectively. The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.