#2473 - Minimum Cost to Buy Apples

Problem Description

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.

You can buy apples in any city you want, but some cities have different costs to buy apples.

You are given the 1-based array appleCost where appleCost[i] is the cost of buying one apple from city i.

You start at some city, traverse through various roads, and eventually buy exactly one apple from any city. After you buy that apple, you have to return back to the city you started at, but now the cost of all the roads will be multiplied by a given factor k.

Given the integer k, return a 1-based array answer of size n where answer[i] is the minimum total cost to buy an apple if you start at city i.

Solution

/**
 * @param {number} n
 * @param {number[][]} roads
 * @param {number[]} appleCost
 * @param {number} k
 * @return {number[]}
 */
var minCost = function(n, roads, appleCost, k) {
  const graph = new Array(n).fill().map(() => []);
  for (const [cityA, cityB, cost] of roads) {
    graph[cityA - 1].push([cityB - 1, cost]);
    graph[cityB - 1].push([cityA - 1, cost]);
  }

  const result = new Array(n);
  for (let startCity = 0; startCity < n; startCity++) {
    result[startCity] = appleCost[startCity];
  }

  const heap = [];
  for (let startCity = 0; startCity < n; startCity++) {
    heap.push([appleCost[startCity], startCity]);
  }
  heap.sort((a, b) => a[0] - b[0]);

  while (heap.length > 0) {
    const [totalCost, currCity] = heap.shift();

    if (result[currCity] < totalCost) continue;

    for (const [neighbor, cost] of graph[currCity]) {
      const newCost = result[currCity] + (k + 1) * cost;
      if (result[neighbor] > newCost) {
        result[neighbor] = newCost;

        let insertIndex = 0;
        while (insertIndex < heap.length && heap[insertIndex][0] <= newCost) {
          insertIndex++;
        }
        heap.splice(insertIndex, 0, [newCost, neighbor]);
      }
    }
  }

  return result;
};