How hard is LeetCode 2093. Minimum Cost to Reach City With Discounts?

LeetCode 2093. Minimum Cost to Reach City With Discounts is rated Medium on LeetCode.

What is the time complexity of LeetCode 2093. Minimum Cost to Reach City With Discounts?

The Python solution on this page runs in O(n log n).

What is the space complexity of LeetCode 2093. Minimum Cost to Reach City With Discounts?

The Python solution on this page uses O(n) auxiliary space.

What topics does LeetCode 2093. Minimum Cost to Reach City With Discounts cover?

LeetCode 2093. Minimum Cost to Reach City With Discounts is tagged Graph, Shortest Path and Heap (Priority Queue) on LeetCode.

Is LeetCode 2093. Minimum Cost to Reach City With Discounts a premium problem?

Yes. LeetCode 2093. Minimum Cost to Reach City With Discounts is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Minimum Cost to Reach City With Discounts — LeetCode 2093 Python Solution

MediumLeetCode PremiumGraphShortest PathHeap (Priority Queue)

Problem: #2093
Pattern: Heap / Priority Queue
Reading time: 4 min
Source: leetcode.com

The problem

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.

Example

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.

Python solution

Python

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

Complexity

Measure	Complexity
Time	O(n log n)
Space	O(n) auxiliary

Pattern: Heap / Priority Queue

Keep only the best k elements, or always pull the smallest, in log time. LeetCode 2093. Minimum Cost to Reach City With Discounts is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Heap (Priority Queue).

The heap / priority queue guide has the Python template for the pattern and the 163 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2093. Minimum Cost to Reach City With Discounts?: LeetCode 2093. Minimum Cost to Reach City With Discounts is rated Medium on LeetCode.
What is the time complexity of LeetCode 2093. Minimum Cost to Reach City With Discounts?: The Python solution on this page runs in O(n log n).
What is the space complexity of LeetCode 2093. Minimum Cost to Reach City With Discounts?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 2093. Minimum Cost to Reach City With Discounts cover?: LeetCode 2093. Minimum Cost to Reach City With Discounts is tagged Graph, Shortest Path and Heap (Priority Queue) on LeetCode.
Is LeetCode 2093. Minimum Cost to Reach City With Discounts a premium problem?: Yes. LeetCode 2093. Minimum Cost to Reach City With Discounts is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.