Finding the percentage in Python HackerRank Solution

Hello Programmers, In this post, you will know how to solve the Finding the percentage in Python HackerRank Solution. This problem is a part of the HackerRank Python Programming Series.

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Finding the percentage in Python HackerRank Solutions

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Finding the percentage in Python HackerRank Solution

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The provided code stub will read in a dictionary containing key/value pairs of name:[marks] for a list of students. Print the average of the marks array for the student name provided, showing 2 places after the decimal.

Example
marks key:value pair are

‘alpha’:[20,30,40]

‘beta’:[30,50,70]

query_name = ‘beta’

The query_name is ‘beta’. beta’s average score is (30+50+70)/3=50.0.

Input Format

The first line contains the integer n, the number of students’ records. The next n lines contain the names and marks obtained by a student, each value separated by a space. The final line contains query_name, the name of a student to query.

Constraints

  • 2 <= N <= 10
  • 0 <= Marks <= 100
  • Length of marks arrays = 3

Output Format

Print one line: The average of the marks obtained by the particular student correct to 2 decimal places.

Sample Input 0

3
Krishna 67 68 69
Arjun 70 98 63
Malika 52 56 60
Malika

Sample Output 0

56.00

Explanation 0

Marks for Malika are  whose average is 

Sample Input 1

2
Harsh 25 26.5 28
Anurag 26 28 30
Harsh

Sample Output 1

26.50

Finding the percentage in Python Hacker Rank Solution

if __name__ == '__main__':
    n = int(input())
    marks = {}
    for _ in range(n):
        name, *line = input().split()
        scores = list(map(float, line))
        scores=sum(scores)/3
        marks[name] = scores
    a = input()
    print('%.2f' % marks[a])

Disclaimer: The above Problem (Finding the percentage) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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