uppose you are going on a road trip with friends. Unfortunately, your headlights are broken, so you can only drive in the daytime. Therefore, on any given day you can drive no more than d miles. You have a map with n different hotels and the distances from your start point to each hotel x 1 < x 2 < ... < x n . Your final destination is the last hotel. Describe an efficient greedy algorithm that determines which hotels you should stay in if you want to minimize the number of days it takes you to get to your destination. What is the running time of your algorithm

Explanation:

Algorithm:

a. In each day, you will have to loop through the hotels that come to the hotel after you stayed last night.

b. If a hotel 'h' is found at more than 'd' distance away from last stayed hotel, then the hotel previous of 'h' is chosen to wait for that night. This is the greedy step, and you stay in this hotel.

c. The process for steps a and b is then repeated until we've reached the last hotel xn.

Running time:

Notice that the worst case occurs if each hotel is at a distance of successive multiples of 'd'. The best move is to estimate the distance to each hotel twice the whole computation in the scenario.

Thus, the total running time that could occur in the worst case is O(2n) = O(n). This is said to be linear time.

so much for the extra points :

answer; peter zenger;