Description and Simple Example
Imagine you have n items of size items[i]
each one. Your goal is to put as many items as possible in the box of size N
, i. e. choose some objects whose summary size is less or equal to size of the box N
.
It's not so hard to understand that you have to choose smaller items first, and you can do it until there is an empty place. If you have items = [2, 1, 5, 5, 3, 7, 6]
and N = 13
, you put items of size 1, 2, 3, 5 and receive 4 items in the box. You can’t put any other item, because left empty space has size 2 and it is not enough. The smaller item you choose every time, the more space is left, and the more other items you can put in.
So, the algorithm is following:
12345678910def maxItems(items, n): items.sort() # reorder items in ascending order ans = 0 for item in items: n -= item # reduce the empty space if n < 0: break ans+=1 # if some empty space left, item is added return ans
Swipe to start coding
Change maxItems()
function to return not the number of items, but the list of items, that should be added to the box.
Løsning
Takk for tilbakemeldingene dine!
single
Spør AI
Spør AI
Spør om hva du vil, eller prøv ett av de foreslåtte spørsmålene for å starte chatten vår
Oppsummer dette kapittelet
Explain code
Explain why doesn't solve task
Awesome!
Completion rate improved to 7.69
Description and Simple Example
Sveip for å vise menyen
Imagine you have n items of size items[i]
each one. Your goal is to put as many items as possible in the box of size N
, i. e. choose some objects whose summary size is less or equal to size of the box N
.
It's not so hard to understand that you have to choose smaller items first, and you can do it until there is an empty place. If you have items = [2, 1, 5, 5, 3, 7, 6]
and N = 13
, you put items of size 1, 2, 3, 5 and receive 4 items in the box. You can’t put any other item, because left empty space has size 2 and it is not enough. The smaller item you choose every time, the more space is left, and the more other items you can put in.
So, the algorithm is following:
12345678910def maxItems(items, n): items.sort() # reorder items in ascending order ans = 0 for item in items: n -= item # reduce the empty space if n < 0: break ans+=1 # if some empty space left, item is added return ans
Swipe to start coding
Change maxItems()
function to return not the number of items, but the list of items, that should be added to the box.
Løsning
Takk for tilbakemeldingene dine!
Awesome!
Completion rate improved to 7.69single