Time Needed To Buy Tickets - Leetcode Solution

The problem "Time Needed To Buy Tickets" on LeetCode can be solved using binary search. The problem statement can be summarized as follows:

There are n bus stations along a route. You want to buy a ticket for all the stations in this route. You can buy tickets only at the bus stations. The cost of a ticket from station i to station j is given as (j - i) * cost[i]. You need to find the minimum cost to buy tickets for all the stations along the route. You can buy tickets at most once at each station.

To solve this problem, we can first find the maximum cost of the entire route, which is (n - 1) * cost[n - 1]. We can use this maximum cost as the upper bound of our binary search. We can set the lower bound as zero.

In each iteration of binary search, we compute the mid cost as the average of the upper and lower bounds. We then use this mid cost to check if it is possible to buy tickets for all the stations along the route within the given mid cost. If it is possible, we update the upper bound as mid cost. If it is not possible, we update the lower bound as mid cost + 1.

To check if it is possible to buy tickets for all the stations along the route within the given mid cost, we can use the following approach:

We can maintain a dp array of size n, where dp[i] represents the minimum cost to buy tickets for all the stations up to station i. We can initialize dp[0] to 0. For each station i, we can calculate dp[i] as follows:

dp[i] = dp[j] + (i - j) * cost[i]

where j is the last station where we bought a ticket on our way to station i. We can loop over all the stations j < i and find the minimum value of dp[j] + (i - j) * cost[i], which gives us the minimum cost to buy tickets for all the stations up to station i.

If dp[n - 1] is less than or equal to the mid cost, it means it is possible to buy tickets for all the stations along the route within the given mid cost. Otherwise, it is not possible.

The time complexity of this solution is O(n log C), where n is the number of stations and C is the maximum cost of the entire route.

Here is the Python code for the solution:

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        n = len(days)
        dp = [float('inf')] * n
        dp[0] = 0
        for i in range(1, n):
            for j in range(i):
                duration = days[i] - days[j] + 1
                if duration <= 1:
                    dp[i] = min(dp[i], dp[j] + costs[0])
                elif duration <= 7:
                    dp[i] = min(dp[i], dp[j] + costs[1])
                else:
                    dp[i] = min(dp[i], dp[j] + costs[2])
        return dp[-1]

In this code, we initialize dp[0] to zero, since we don't have to buy a ticket to go from the starting station to the first station. For each station i, we loop over all the stations j < i and calculate the minimum cost to buy tickets for all the stations up to i. Finally, we return dp[-1], which represents the minimum cost to buy tickets for all the stations along the route.

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