🐚 [Exercise] Collections - Need help 💫 Python 🐍

Question

Hello, I have some trouble with an exercise at school. Was wondering if you could help me solving this, and if possible explaining it as well ? We haven't really seen the Collections yet...

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The following code sample is a naive implementation of the ​intersection function, where the intersection of two collections, A and B, is defined as a collection that contains all the elements that are common to both A and B.

def​ intersection​(​A​,​ B​):
intersection ​=​ ​[]
for​ a ​in​ A​:
​ if​ a ​in​ B​:
intersection​.​append​(a​ ​)
​ return​ intersection

Write code for each function listed below (union and difference), where:

🌟 The ​union ​of two collections, A and B, is defined as a collection that contains all the elements that are in either A or B

🌟 The ​difference ​of two collections, A and B, is defined as a collection that contains the elements that are in either A or B, but ​not​ both.

def union(A, B):
...

def difference(A, B):
...

Answer 1

Where is your try?

Answer 2

Okay Jin 🌙 .
For the & operator, you can iterate through A to see if it is in B. Then, remove 1 copy of that element in both A and B. Repeat until completion.

For the difference, you need to combine both A and B, remove A&B twice.

Answer 3

Jin 🌙
You tried.
This is a not so naive implementation.

https://code.sololearn.com/crUR9AY4tR8x/?ref=app

Answer 4

Jin 🌙 Are the elements in the collection unique? If they are, Python has a unique solution: sets.

Correct me if the collections have repeated elements.

Answer 5

Prometheus:
In the exercise, I understood that they have common elements, so not unique then ? I believe they have repeated one.
Are the Collections like the Array in C ?

KrOW: I need to understand the exercise before attempt it.

Answer 6

I think that the exercises are only for make you thinking about the algorithm assuming than a collection is a simple list.

Answer 7

My attempt is below, hope it makes sense ?

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def union(A, B):
union= []
for a in A:
union.append(a)
for b in B:
union.append(b)
return union

def difference(A, B):
difference = []
Union= union(A,B)
Intersection= intersection(A,B)

for u in Union
if not u in Intersection
difference.append(u)
return difference

Answer 8

The union is not right... You add to it every item of either collections but wouldnt be the case... At example:
A=[1, 2, 3]
B= [3,4,5]

Your code return like union:
[1,2,3,3,4,5] but it would return [1,2,3,4,5]
This is because you have to make sure to include into union only items not yet included