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In this post, you will find the solution for **set.union() Operators in Python-HackerRank Problem**. We are providing the **correct and tested solutions** of coding problems present on **HackerRank**. If you are not able to solve any problem, then you can take help from our Blog/website.

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**Introduction To Python**

**Python **is a widely-used, interpreted, **object-oriented, and high-level programming** language with dynamic semantics, used for general-purpose programming. It was created by **Guido van Rossum**, and first released on **February 20, 1991**.

**Python **is a computer programming language often used to build websites and software, automate tasks, and conduct data analysis. It is also used to create various machine learning algorithm, and helps in Artificial Intelligence. Python is a general purpose language, meaning it can be used to create a variety of different programs and isn’t specialized for any specific problems. This versatility, along with its beginner-friendliness, has made it one of the most-used programming languages today. A survey conducted by industry analyst firm **RedMonk **found that it was the most popular programming language among developers in **2020**.

** Link for the Problem** – set.union() Operators in Python – HackerRank Solution

set.union() Operators in Python – HackerRank Solution

**.union()**

The *.union()* operator returns the union of a set and the set of elements in an iterable.

Sometimes, the *|* operator is used in place of *.union()* operator, but it operates only on the set of elements in *set*.

Set is immutable to the *.union()* operation (or *|* operation).

**Example**

>>> s = set("Hacker") >>> print s.union("Rank") set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n']) >>> print s.union(set(['R', 'a', 'n', 'k'])) set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n']) >>> print s.union(['R', 'a', 'n', 'k']) set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n']) >>> print s.union(enumerate(['R', 'a', 'n', 'k'])) set(['a', 'c', 'r', 'e', (1, 'a'), (2, 'n'), 'H', 'k', (3, 'k'), (0, 'R')]) >>> print s.union({"Rank":1}) set(['a', 'c', 'r', 'e', 'H', 'k', 'Rank']) >>> s | set("Rank") set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

**Task**

The students of District College have subscriptions to *English* and *French* newspapers. Some students have subscribed only to *English*, some have subscribed to only *French* 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 the other set is subscribed to the *French* newspaper. The same student could be in both sets. Your task is to find the total number of students who have subscribed to *at least one* newspaper.

**Input Format**

The first line contains an integer, , the number of students who have subscribed to the *English* newspaper.

The second line contains space separated roll numbers of those students.

The third line contains , the number of students who have subscribed to the *French* newspaper.

The fourth line contains space separated roll numbers of those students.

**Constraints**

**Output Format**

Output the total number of students who have *at least one* subscription.

**Sample Input**

9 1 2 3 4 5 6 7 8 9 9 10 1 2 3 11 21 55 6 8

**Sample Output**

13

**Explanation**

Roll numbers of students who have *at least one* subscription:

and . Roll numbers: and are in both sets so they are only counted once.

Hence, the total is students.

set.union() Operators in Python – HackerRank Solution

# set.union() Operators in Python - Hacker Rank Solution # Python 3 # Enter your code here. Read input from STDIN. Print output to STDOUT # set.union() Operators in Python - Hacker Rank Solution START N1 = int(input()) storage1 = set(input().split()); N2 = int(input()) storage2 = set(input().split()); storage3 = storage1.union(storage2) print(len(storage3)) # set.union() Operators in Python - Hacker Rank Solution END

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