Dr. Mickhead's family is visiting him next Sunday to have dinner together. He wishes the party to be really special, so he plans to use his best plates for the table setting. Dr. Mickhead has N stacks of plates. Each stack contains K plates. Each plate has a positive beauty value, describing how beautiful it looks. Dr. Mickhead would like to take exactly P plates to use for dinner. If he would like to take a plate from a stack, he must also take all of the plates above it. Help Dr. Mickhead to calculate the maximum total sum of beauty values that can be achieved by picking P plates.
Note: P is always less than or equal to the total amount of plates in all stacks.
This mission was proposed by Ben Jacobson and Wajeb Saab for Google Kickstart Round A 2020.
Input: Two arguments
- The first one is a list of lists of equal size. The i-th list contains positive integers, describing the beauty values of i-th stack of plates from top to bottom;
- The second is a positive integer. The amount of plates Dr. Mickhead is willing to put on the table.
Output: An integer - the maximum total sum of beauty values that can be achieved with respect to the constraints.
Consider the following input: [[10, 10, 100, 30], [80, 50, 10, 50,]], 5 . So the best option would be picking 3 plates from the first stack and 2 plates from the second stack. In total, the sum of beauty values is 250.
[10, 10, 100, _] -> total: 10 + 10 + 100 = 120 [80, 50, _, _] -> total: 80 + 50 = 130 ---------------------------------------------- total: 120 + 130 = 250
best_plates([[10, 10, 100, 30], [80, 50, 10, 50,]], 5) == 250 best_plates([[80, 80], [15, 50], [20, 10]], 3) == 180
How it’s used: This is a variation of a very well-known knapsack problem.
Precondition: 1 ≤ len(stacks) ≤ 100 1 ≤ len(stacks[i]) ≤ 50 1 ≤ stacks[i][j] ≤ 200
CheckiO Extensions allow you to use local files to solve missions. More info in a blog post.
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pip3 install checkio_client
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checkio install-plugin --chromium