Question

2093. Minimum Cost to Reach City With Discounts

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Description

A series of highways connect n cities numbered from 0 to n - 1. You are given a 2D integer array highways where highways[i] = [city1i, city2i, tolli] indicates that there is a highway that connects city1i and city2i, allowing a car to go from city1i to city2i and vice versa for a cost of tolli.

You are also given an integer discounts which represents the number of discounts you have. You can use a discount to travel across the ith highway for a cost of tolli / 2 (integer division). Each discount may only be used once, and you can only use at most one discount per highway.

Return _the minimum total cost to go from city _0_ to city _n - 1_, or _-1_ if it is not possible to go from city _0_ to city _n - 1_._

 

Example 1:

Input: n = 5, highways = [[0,1,4],[2,1,3],[1,4,11],[3,2,3],[3,4,2]], discounts = 1
Output: 9
Explanation:
Go from 0 to 1 for a cost of 4.
Go from 1 to 4 and use a discount for a cost of 11 / 2 = 5.
The minimum cost to go from 0 to 4 is 4 + 5 = 9.

Example 2:

Input: n = 4, highways = [[1,3,17],[1,2,7],[3,2,5],[0,1,6],[3,0,20]], discounts = 20
Output: 8
Explanation:
Go from 0 to 1 and use a discount for a cost of 6 / 2 = 3.
Go from 1 to 2 and use a discount for a cost of 7 / 2 = 3.
Go from 2 to 3 and use a discount for a cost of 5 / 2 = 2.
The minimum cost to go from 0 to 3 is 3 + 3 + 2 = 8.

Example 3:

Input: n = 4, highways = [[0,1,3],[2,3,2]], discounts = 0
Output: -1
Explanation:
It is impossible to go from 0 to 3 so return -1.

 

Constraints:

• 2 <= n <= 1000
• 1 <= highways.length <= 1000
• highways[i].length == 3
• 0 <= city1i, city2i <= n - 1
• city1i != city2i
• 0 <= tolli <= 105
• 0 <= discounts <= 500
• There are no duplicate highways.

Solutions

Solution 1

class Solution:
  def minimumCost(self, n: int, highways: List[List[int]], discounts: int) -> int:
    g = defaultdict(list)
    for a, b, c in highways:
      g[a].append((b, c))
      g[b].append((a, c))
    q = [(0, 0, 0)]
    dist = [[inf] * (discounts + 1) for _ in range(n)]
    while q:
      cost, i, k = heappop(q)
      if k > discounts:
        continue
      if i == n - 1:
        return cost
      if dist[i][k] > cost:
        dist[i][k] = cost
        for j, v in g[i]:
          heappush(q, (cost + v, j, k))
          heappush(q, (cost + v // 2, j, k + 1))
    return -1

class Solution {
  public int minimumCost(int n, int[][] highways, int discounts) {
    List<int[]>[] g = new List[n];
    for (int i = 0; i < n; ++i) {
      g[i] = new ArrayList<>();
    }
    for (var e : highways) {
      int a = e[0], b = e[1], c = e[2];
      g[a].add(new int[] {b, c});
      g[b].add(new int[] {a, c});
    }
    PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
    q.offer(new int[] {0, 0, 0});
    int[][] dist = new int[n][discounts + 1];
    for (var e : dist) {
      Arrays.fill(e, Integer.MAX_VALUE);
    }
    while (!q.isEmpty()) {
      var p = q.poll();
      int cost = p[0], i = p[1], k = p[2];
      if (k > discounts || dist[i][k] <= cost) {
        continue;
      }
      if (i == n - 1) {
        return cost;
      }
      dist[i][k] = cost;
      for (int[] nxt : g[i]) {
        int j = nxt[0], v = nxt[1];
        q.offer(new int[] {cost + v, j, k});
        q.offer(new int[] {cost + v / 2, j, k + 1});
      }
    }
    return -1;
  }
}

class Solution {
public:
  int minimumCost(int n, vector<vector<int>>& highways, int discounts) {
    vector<vector<pair<int, int>>> g(n);
    for (auto& e : highways) {
      int a = e[0], b = e[1], c = e[2];
      g[a].push_back({b, c});
      g[b].push_back({a, c});
    }
    priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> q;
    q.push({0, 0, 0});
    vector<vector<int>> dist(n, vector<int>(discounts + 1, INT_MAX));
    while (!q.empty()) {
      auto [cost, i, k] = q.top();
      q.pop();
      if (k > discounts || dist[i][k] <= cost) continue;
      if (i == n - 1) return cost;
      dist[i][k] = cost;
      for (auto [j, v] : g[i]) {
        q.push({cost + v, j, k});
        q.push({cost + v / 2, j, k + 1});
      }
    }
    return -1;
  }
};