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Impara Description and Simple Example | Greedy Algorithms: Overview and Examples
Greedy Algorithms using Python

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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:

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def 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
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Compito

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.

Soluzione

Switch to desktopCambia al desktop per esercitarti nel mondo realeContinua da dove ti trovi utilizzando una delle opzioni seguenti
Tutto è chiaro?

Come possiamo migliorarlo?

Grazie per i tuoi commenti!

Sezione 1. Capitolo 2
Siamo spiacenti che qualcosa sia andato storto. Cosa è successo?

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book
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:

12345678910
def 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
copy
Compito

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.

Soluzione

Switch to desktopCambia al desktop per esercitarti nel mondo realeContinua da dove ti trovi utilizzando una delle opzioni seguenti
Tutto è chiaro?

Come possiamo migliorarlo?

Grazie per i tuoi commenti!

Sezione 1. Capitolo 2
Switch to desktopCambia al desktop per esercitarti nel mondo realeContinua da dove ti trovi utilizzando una delle opzioni seguenti
Siamo spiacenti che qualcosa sia andato storto. Cosa è successo?
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