LeetCode 2473 - Minimum Cost to Buy Apples

LeetCode Problem 2473

Difficulty: 🟡 Medium
Topics: Array, Graph Theory, Heap (Priority Queue), Shortest Path

Solution

Problem Understanding

The problem asks us to determine the minimum cost to buy exactly one apple starting from each city in a network of cities connected by bidirectional roads. Each city has a specific cost for buying an apple, and each road has a travel cost. After buying an apple, you must return to the starting city, but on the return trip, road costs are multiplied by a given factor k.

The input consists of the number of cities n, a list of roads where each road is represented by [ai, bi, costi], a list appleCost of size n representing the cost of an apple in each city, and a multiplier k. The output should be a list of size n where each entry represents the minimum total cost to buy one apple starting from that city.

Constraints indicate that the number of cities can be up to 1000 and roads up to 2000, which implies that a naive solution that checks every path from every city to every other city would be too slow. There are no repeated edges, and all costs are positive integers, so we do not have to handle zero-cost cycles.

Edge cases include scenarios where it is optimal to buy the apple in the starting city (e.g., when appleCost[start] is smaller than traveling to other cities) or where k = 1 so the return trip cost is the same as the outbound trip. Another subtle case occurs when some cities are isolated but still reachable through multiple roads, ensuring that Dijkstra's algorithm can handle disconnected graphs gracefully.

Approaches

Brute Force

The brute-force approach is to compute the cost of buying an apple at every possible city for each starting city. For a start city i, we would enumerate all possible destinations j, compute the shortest path to j, buy the apple, and then compute the shortest path back to i multiplied by k. The minimum across all possible j is the answer for start city i. This approach requires running a shortest path algorithm (like Dijkstra) n times for each starting city and destination pair, leading to O(n^3) time complexity, which is too slow for the constraints.

Optimal Approach

The key observation is that the cost of traveling to a city and back is symmetric except for the multiplier k on the return trip. By modeling the problem as a weighted graph and using Dijkstra's algorithm from each city to compute the minimum cost to all other cities, we can compute the total cost for each start city efficiently. We optimize further by treating the apple purchase as an additional node with a cost equal to appleCost[j]. This allows us to combine travel and purchase costs in a single pass of Dijkstra for each city, resulting in O(n (m log n)) complexity, which is feasible given the constraints (m is number of edges).

Algorithm Walkthrough

1. Graph Construction: Build an adjacency list for the cities using the given roads. Each city points to a list of tuples (neighbor, cost).
2. Iterate Over Cities: For each starting city i, we will compute the minimum cost to buy an apple.
3. Dijkstra Initialization: For the starting city, initialize a min-heap priority queue with (currentCost, city) = (0, start).
4. Track Minimum Costs: Maintain a dist array that stores the shortest distance from start city to every other city.
5. Process Heap: Pop the city with the smallest cost from the heap, and for each neighbor, check if traveling through the current city offers a shorter distance. If yes, update dist[neighbor] and push it to the heap.
6. Compute Total Cost: After obtaining shortest paths, compute the total cost to buy an apple at each city j as dist[j] + appleCost[j] + k * dist[j]. Keep the minimum across all j for the answer.
7. Repeat for All Cities: Repeat the Dijkstra traversal for all cities 1..n and store the results.

Why it works: Dijkstra's algorithm guarantees the shortest path in graphs with positive edge weights. By multiplying the shortest path distance by k for the return trip, we correctly account for the asymmetric cost. Adding appleCost[j] ensures that the algorithm considers the actual purchase cost, and taking the minimum over all cities guarantees we find the optimal solution.

Python Solution

from heapq import heappush, heappop
from typing import List

class Solution:
    def minCost(self, n: int, roads: List[List[int]], appleCost: List[int], k: int) -> List[int]:
        graph = [[] for _ in range(n)]
        for u, v, w in roads:
            graph[u-1].append((v-1, w))
            graph[v-1].append((u-1, w))
        
        answer = [0] * n
        
        for start in range(n):
            dist = [float('inf')] * n
            dist[start] = 0
            heap = [(0, start)]
            
            while heap:
                d, node = heappop(heap)
                if d > dist[node]:
                    continue
                for nei, w in graph[node]:
                    if dist[node] + w < dist[nei]:
                        dist[nei] = dist[node] + w
                        heappush(heap, (dist[nei], nei))
            
            min_cost = float('inf')
            for j in range(n):
                total = dist[j] + appleCost[j] + k * dist[j]
                min_cost = min(min_cost, total)
            answer[start] = min_cost
        
        return answer

The Python solution first constructs the graph as an adjacency list for efficient traversal. Dijkstra’s algorithm computes the minimum travel cost from the starting city to all other cities. After computing the shortest paths, the total cost to buy an apple at each city, including the return trip cost, is calculated and the minimum is stored.

Go Solution

package main

import (
	"container/heap"
	"math"
)

type Item struct {
	cost int
	node int
}

type PriorityQueue []Item

func (pq PriorityQueue) Len() int { return len(pq) }
func (pq PriorityQueue) Less(i, j int) bool { return pq[i].cost < pq[j].cost }
func (pq PriorityQueue) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }

func (pq *PriorityQueue) Push(x any) { *pq = append(*pq, x.(Item)) }
func (pq *PriorityQueue) Pop() any {
	old := *pq
	n := len(old)
	item := old[n-1]
	*pq = old[:n-1]
	return item
}

func minCost(n int, roads [][]int, appleCost []int, k int) []int64 {
	graph := make([][][2]int, n)
	for _, r := range roads {
		u, v, w := r[0]-1, r[1]-1, r[2]
		graph[u] = append(graph[u], [2]int{v, w})
		graph[v] = append(graph[v], [2]int{u, w})
	}
	
	answer := make([]int64, n)
	for start := 0; start < n; start++ {
		dist := make([]int, n)
		for i := range dist {
			dist[i] = math.MaxInt32
		}
		dist[start] = 0
		pq := &PriorityQueue{}
		heap.Init(pq)
		heap.Push(pq, Item{0, start})
		
		for pq.Len() > 0 {
			item := heap.Pop(pq).(Item)
			d, node := item.cost, item.node
			if d > dist[node] { continue }
			for _, nei := range graph[node] {
				if dist[node] + nei[1] < dist[nei[0]] {
					dist[nei[0]] = dist[node] + nei[1]
					heap.Push(pq, Item{dist[nei[0]], nei[0]})
				}
			}
		}
		
		minCost := int64(math.MaxInt64)
		for j := 0; j < n; j++ {
			total := int64(dist[j]) + int64(appleCost[j]) + int64(k)*int64(dist[j])
			if total < minCost {
				minCost = total
			}
		}
		answer[start] = minCost
	}
	
	return answer
}

The Go solution mirrors the Python version, using a priority queue from container/heap to implement Dijkstra's algorithm. We carefully use int64 for results to avoid overflow since costs and multipliers can be large.

Worked Examples

Example 1:

Starting at city 1:

• Distances after Dijkstra: [0,4,4,5] (shortest paths to cities 1-4)

• Total cost calculation:

• City 1: 0 + 56 + 0*2 = 56

• City 2: 4 + 42 + 4*