Introduction

In this blog post, we will explore how to solve the LeetCode problem "1928" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is a country of n cities numbered from 0 to n - 1 where all the cities are connected by bi-directional roads. The roads are represented as a 2D integer array edges where edges[i] = [xi, yi, timei] denotes a road between cities xi and yi that takes timei minutes to travel. There may be multiple roads of differing travel times connecting the same two cities, but no road connects a city to itself. Each time you pass through a city, you must pay a passing fee. This is represented as a 0-indexed integer array passingFees of length n where passingFees[j] is the amount of dollars you must pay when you pass through city j. In the beginning, you are at city 0 and want to reach city n - 1 in maxTime minutes or less. The cost of your journey is the summation of passing fees for each city that you passed through at some moment of your journey (including the source and destination cities). Given maxTime, edges, and passingFees, return the minimum cost to complete your journey, or -1 if you cannot complete it within maxTime minutes. Example 1: Input: maxTime = 30, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3] Output: 11 Explanation: The path to take is 0 -> 1 -> 2 -> 5, which takes 30 minutes and has $11 worth of passing fees. Example 2: Input: maxTime = 29, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3] Output: 48 Explanation: The path to take is 0 -> 3 -> 4 -> 5, which takes 26 minutes and has $48 worth of passing fees. You cannot take path 0 -> 1 -> 2 -> 5 since it would take too long. Example 3: Input: maxTime = 25, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3] Output: -1 Explanation: There is no way to reach city 5 from city 0 within 25 minutes. Constraints: 1 <= maxTime <= 1000 n == passingFees.length 2 <= n <= 1000 n - 1 <= edges.length <= 1000 0 <= xi, yi <= n - 1 1 <= timei <= 1000 1 <= passingFees[j] <= 1000 The graph may contain multiple edges between two nodes. The graph does not contain self loops.

Explanation

Here's the solution to the problem, focusing on efficiency and optimality:

• High-Level Approach:

  • Use Dynamic Programming to store the minimum cost to reach each city within a given time limit. dp[i][t] stores the minimum cost to reach city i within time t.
  • Iterate through all possible times from 1 to maxTime. For each time t, update the dp table by considering all possible edges and the corresponding time and cost to reach the connected cities.
  • The final answer is stored in dp[n-1][maxTime]. Return -1 if the value is infinity, which implies no path exists within the given time limit.

• Complexity:

  • Runtime: O(maxTime * E), where E is the number of edges. The complexity comes from iterating through the maxTime and then edges from each node.
  • Storage: O(n * maxTime) where n is the number of cities. We store the minimum costs for each city at different times.

Code

    import sys

def minCost(maxTime: int, edges: list[list[int]], passingFees: list[int]) -> int:
    """
    Calculates the minimum cost to reach city n-1 from city 0 within maxTime,
    considering passing fees and travel times.
    """
    n = len(passingFees)

    # dp[i][t] is the minimum cost to reach city i within time t
    dp = [[float('inf')] * (maxTime + 1) for _ in range(n)]

    # Initialize the cost to reach city 0 at time 0
    dp[0][0] = passingFees[0]

    # Iterate through all possible times
    for t in range(1, maxTime + 1):
        # Iterate through all edges
        for u, v, time in edges:
            # If we can reach city u within time t - time, update the cost to reach city v
            if t >= time and dp[u][t - time] != float('inf'):
                dp[v][t] = min(dp[v][t], dp[u][t - time] + passingFees[v])

            # Also, consider the reverse direction (since edges are bidirectional)
            if t >= time and dp[v][t - time] != float('inf'):
                dp[u][t] = min(dp[u][t], dp[v][t - time] + passingFees[u])

    # Find the minimum cost to reach city n-1 within any time <= maxTime
    min_cost_to_target = min(dp[n - 1])

    return min_cost_to_target if min_cost_to_target != float('inf') else -1