Set.difference() Operation


Problem Statement :


.difference()
The tool .difference() returns a set with all the elements from the set that are not in an iterable.
Sometimes the - operator is used in place of the .difference() tool, but it only operates on the set of elements in set.
Set is immutable to the .difference() operation (or the - operation).

>>> s = set("Hacker")
>>> print s.difference("Rank")
set(['c', 'r', 'e', 'H'])

>>> print s.difference(set(['R', 'a', 'n', 'k']))
set(['c', 'r', 'e', 'H'])

>>> print s.difference(['R', 'a', 'n', 'k'])
set(['c', 'r', 'e', 'H'])

>>> print s.difference(enumerate(['R', 'a', 'n', 'k']))
set(['a', 'c', 'r', 'e', 'H', 'k'])

>>> print s.difference({"Rank":1})
set(['a', 'c', 'e', 'H', 'k', 'r'])

>>> s - set("Rank")
set(['H', 'c', 'r', 'e'])
Task
Students of District College have a subscription to English and French newspapers. Some students have subscribed to only the English newspaper, some have subscribed to only the French newspaper, and some have subscribed to both newspapers.

You are given two sets of student roll numbers. One set has subscribed to the English newspaper, and one set has subscribed to the French newspaper. Your task is to find the total number of students who have subscribed to only English newspapers.


Input Format

The first line contains the number of students who have subscribed to the English newspaper.
The second line contains the space separated list of student roll numbers who have subscribed to the English newspaper.
The third line contains the number of students who have subscribed to the French newspaper.
The fourth line contains the space separated list of student roll numbers who have subscribed to the French newspaper.


Constraints
0 < total number of students in the college <1000


Output Format

Output the total number of students who are subscribed to the English newspaper only.


Solution :



title-img 


                            Solution in C :

num1 = int(input())
setA = set(map(int, input().split()))
num2 = int(input())
setB = set(map(int, input().split()))

print(len(setA.difference(setB)))








View More Similar Problems

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →