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Sum of Odd Cubed Numbers in Python

The problem

Discover the sum of the odd numbers inside an array, after cubing the preliminary integers. The operate ought to return None if any of the values aren’t numbers.

Be aware: Booleans shouldn’t be thought of as numbers.

The answer in Python code

Possibility 1:

def cube_odd(arr):
    if any(kind(x) is just not int for x in arr):
        return None
    return sum(x ** 3 for x in arr if x % 2 != 0)

Possibility 2:

def cube_odd(arr):
    if len(set(map(kind,arr))) < 2:
        return sum(n**3 for n in arr if npercent2)

Possibility 3:

def cube_odd(arr):
    s = 0
    for n in arr:
        if kind(n) != int: break
        elif npercent2: s += n**3
        return s

Check instances to validate our resolution

import take a look at
from resolution import cube_odd

@take a look at.describe("Mounted Assessments")
def fixed_tests():
    @take a look'Primary Check Instances')
    def basic_test_cases():
        take a look at.assert_equals(cube_odd([1, 2, 3, 4]), 28)
        take a look at.assert_equals(cube_odd([-3,-2,2,3]), 0)
        take a look at.assert_equals(cube_odd(["a",12,9,"z",42]), None)
        take a look at.assert_equals(cube_odd([True,False,2,4,1]), None)
@take a look at.describe("Random checks")
def _():
    from random import randint, random
    sol=lambda arr: None if any(kind(e)!=int for e in arr) else sum(e*e*e for e in arr if epercent2)
    for _ in vary(40):
        arr=[randint(-10,10) for q in range(randint(5,10))]
        arr=arr if randint(0,1) else sorted(arr+[base[randint(0,len(base)-1)] for q in vary(randint(5,10))],key=lambda a: random())
        anticipated = sol(arr)
        @take a look"Testing for cube_odd({arr})")
        def _():
            take a look at.assert_equals(cube_odd(arr[:]),anticipated)


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