The key idea of the greedy algorithm for Garfield's problem is to sort the TV shows based on their ending times in ascending order. We will then keep track of the latest ending time among the shows that Garfield has selected so far. We will iterate through the sorted list of TV shows, and for each show, if its starting time is after the latest ending time, we will select that show and update the latest ending time accordingly. By selecting the shows in this manner, we ensure that Garfield is always watching a show that ends as late as possible, and we minimize the number of shows he has to switch between.
To implement this algorithm in O(n log n) time, we can first sort the list of TV shows based on their ending times, which takes O(n log n) time. We can then iterate through the sorted list once to select the shows, which takes O(n) time. Therefore, the overall time complexity of the algorithm is O(n log n).
To prove the correctness of the algorithm, we can use a proof by contradiction. Suppose that there exists a smaller set of TV shows that Garfield can watch to occupy the entire time period. Let this set of shows be S', and let the last show in S' end at time t. Since S' is a smaller set, there must exist a show in the sorted list that ends after t. However, since we selected the shows in the sorted list based on their ending times, this show must also be in Garfield's set of selected shows. Therefore, Garfield's set of selected shows is at least as small as S', and the algorithm is correct.
Overall, the greedy algorithm described above is an efficient O(n log n) solution to Garfield's problem, and it is guaranteed to give the optimal solution.