Solved Homework: Question: Little Johnny is extremely fond of watching television. His parents are off for work for the peri…

Little Johnny is extremely fond of watching television. His parents are off for work for the period [S, F), and he wants to make full use of this time by watching as much television as possible: in fact, he wants to watch TV non-stop the entire period [S, F). He has a list of his favorite n TV shows (on different channels), where the i-th show runs for the time period [sifi), so that the union of [si,fi) fully covers the entire time period IS, F) when his parents are away. Little Johnny doesnt mind switching in the middle of a show he is watching, but is very lazy to switch TV channels, so he wants to find the smallest set of TV shows that he can watch, and still stay occupied for the entire period [S, F). Design an efficient O(n logn) greedy algorithm to help Little Johny.

Little Johnny is extremely fond of watching television. His parents are off for work for the period [S, F), and he wants to make full use of this time by watching as much television as possible: in fact, he wants to watch TV non-stop the entire period [S, F). He has a list of his favorite n TV shows (on different channels), where the i-th show runs for the time period [sifi), so that the union of [si,fi) fully covers the entire time period IS, F) when his parents are away. Little Johnny doesn’t mind switching in the middle of a show he is watching, but is very lazy to switch TV channels, so he wants to find the smallest set of TV shows that he can watch, and still stay occupied for the entire period [S, F). Design an efficient O(n logn) greedy algorithm to help Little Johny.

Expert Answer

answers

Greedy Alogrithm for the above problem.

  1. Store the start time (si) and end time (fi) in a vector of pairs.
  2. Now make a copy of the vector formed and sort the first in increasing order of starting time and the other in increasing order of end time.
  3. Now once iterate the vector and find out the show which has the largest duration.
  4. Now iterate towards left in the start time vector and take the lowest value whose ending time is greaterthan or equal to current shows starting time and include that show in the set.
  5. Similarly iterate towards right in the end time vector and take the largest value whose starting time is lessthan or equal to current shows end time.
  6. Repeat the steps 4 and 5 until we reach S in start time vector and F in end time vector.

The algorithm has average run time compolexity of O(n log(n)) due to sorting of vectors.

If any doubt then comment. Hope it helps 😀

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