set in python | Lec-17

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In this video, I have talked about the fundamentals of programming and Python. If you complete this, you will be well-versed in using tuples for different use cases.
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Опубликовано:

29 авг 2024

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Комментарии : 18

@BeingSam7 16 дней назад

Manish, humne to padha hai only strings and tuples are immutable. list, set and dictionary are mutable bro..

@manish_kumar_1 16 дней назад

Hi everyone, Yes @Beingsam is correct here. We can add and remove elements from a set so it is mutable. Since this is unordered so, items of a set are immutable but set is mutable. For example set1 = {1,4,7,3,2,8} I can add the element 12 in the set so it is mutable. But if I want to replace 2 with 12 so it is not possible because as I said earlier elements of a set are immutable. I am really sorry for my mistake 😢

@BeingSam7 16 дней назад

We all are humans, and I have said it earlier you are a great human being. It takes a lot of courage to admit of being wrong and pin the very comment saying so. You are the best teacher we could have asked for, be it Pyspark or python. You Rock @Manish_kumar_1

@jigsparikh7961 4 месяца назад

i always shyed away from python but because of you i started learning it. Thanks a million you are awesome.

@ShubhamKumar-cf9ih 4 месяца назад

Time and Space both will be O(n). As it has to traverse to each number and for storing the number also it has to add all till the match.

@raghavjuyalcomedyclub184 3 месяца назад

time complexity is o(n^3) since to ( first loop o(n) x check element in list1 o(n) x check in list2 o(n) )

@user-yd1fu7hl6j 4 месяца назад

time complexity - o(n) becoz of only one loop , extra spacis used but in worst case maximum element can be appended then o(n) space complexity

@anantkashyap1746 4 месяца назад

Time complexity O(n) as it traverse using for loop and space complexity O(N) as it uses space to append the result

@jigsparikh7961 4 месяца назад

superb video thanks

@satyamkumarjha4185 Месяц назад

sets are mutable.

@safiadawood5637 4 месяца назад

# Q1 list1 = [1,2,3,4,5,6] list2 = [4,5,6,7,8] t1 = set(list1) t2 = set(list2) print(f"Missing values in list1 ={list(t2.difference(t1))}") print(f"Additional values in list2 = {list(t1.difference(t2))}") print(f"Missing values in list1 = {list (t1.difference(t2))}") print(f"Additional values in list2 ={list(t2.difference(t1))}") # Q2 ar1 = [1, 5, 10, 20, 40, 80] ar2 = [6, 7, 20, 80, 100] ar3 = [3, 4, 15, 20, 30, 70, 80, 120] common_items = set(ar1).intersection(set(ar2),set(ar3)) print(list(common_items)) Time complexity will be O(n) to convert to set, and O(n) to do the intersetion. So , final time complexity will be O(n) Space complexity will be O(n) too as each list will be stored as set.

@bolisettisaisatwik2198 4 месяца назад

Big fan Manish Kumar :).

@shoaibmalik1217 4 месяца назад

First comment

@PranshuHasani 4 месяца назад

@AbhinavShukla-ki5jd Месяц назад

#Q1 from loguru import logger l1=[1,2,3,4,5,6] s1=set(l1) l2=[4,5,6,7,8] s2=set(l2) logger.info(f"Missing values in list1 ={list(s2-s1)}") logger.info(f"Additional values in list1 ={list(s1-s2)}") logger.info(f"Missing values in list2 ={list(s1-s2)}") logger.info(f"Additional values in list1 ={list(s2-s1)}") #Q2 l1=[1,5,10,20,40,80] l2=[6,7,20,80,100] l3=[3,4,15,20,30,70,80,100] s1=set(l1) s2=set(l2) s3=set(l3) sn=s1.intersection(s2,s3) logger.info(list(sn))