Maximize It! in Python | HackerRank Solution

Hello coders, today we are going to solve Maximize It! HackerRank Solution in Python.

Maximize It! in Python

Contents

Task

You are given a function f(X) = X2. You are also given K lists. The ith list consists of Ni elements.

You have to pick one element from each list so that the value from the equation below is maximized:

S = (f(X1) + f(X2) + . . . + f(Xk) % M

Xi denotes the element picked from the ith list . Find the maximized value Smax obtained.

% denotes the modulo operator.

Note that you need to take exactly one element from each list, not necessarily the largest element. You add the squares of the chosen elements and perform the modulo operation. The maximum value that you can obtain, will be the answer to the problem.

Input Format

The first line contains 2 space separated integers K and M.
The next K lines each contains an integer Ni, denoting the number of elements in the ith list, followed by Ni space separated integers denoting the elements in the list.

Constraints

  • 1 <= K <= 7
  • 1 <= M <= 1000
  • 1 <= Ni <= 7
  • 1 <= Magnitude of elements in list <= 109

Output Format

Output a single integer denoting the value Smax.

Sample Input

3 1000
2 5 4
3 7 8 9 
5 5 7 8 9 10 

Sample Output

206

Solution – Maximize It! in Python

# Enter your code here. Read input from STDIN. Print output to STDOUT
import itertools

NUMBER_OF_LISTS, MODULUS = map(int, input().split())
LISTS_OF_LISTS = []

for i in range(0, NUMBER_OF_LISTS):
    new_list = list(map(int, input().split()))
    del new_list[0]
    LISTS_OF_LISTS.append(new_list)

def squared(element):
    return element**2

COMBS = list(itertools.product(*LISTS_OF_LISTS))
RESULTS = []

for i in COMBS:
    result1 = sum(map(squared, [a for a in i]))
    result2 = result1 % MODULUS
    RESULTS.append(result2)

print(max(RESULTS))

Disclaimer: The above Problem (Maximize It!) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

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