Final answer:
To find an optimal solution for this problem, we can use a greedy algorithm. We start by placing the first song on the first CD and then iterate through the remaining songs, checking if they can fit on the current CD. The algorithm guarantees the correct placement of songs and minimizes the number of CDs needed.
Step-by-step explanation:
To find an optimal solution for this problem, we can use a greedy algorithm. We can start by placing the first song on the first CD, and then iterate through the remaining songs. For each song, we check if it can fit on the current CD. If it can, we add the song to the CD. If it can't, we start a new CD and add the song to it. This process continues until all songs have been placed on CDs.
Here is the algorithm:
- Create an empty CD list.
- Create a new CD and add the first song to it.
- Iterate through the remaining songs:
- If the current song can fit on the current CD, add it to the CD.
- If the current song can't fit on the current CD, create a new CD and add the song to it.
Return the number of CDs created.
This algorithm is correct because it guarantees that all songs will be placed on CDs in the given order, with no song split across CDs. It also minimizes the number of CDs needed, as it tries to fit as many songs as possible on each CD before moving on to a new CD.
The time complexity of this algorithm is O(n), where n is the number of songs. This is because we iterate through each song once to determine which CD it should be placed on. The other operations, such as creating CDs and adding songs, have constant time complexity.