Sum of Odd Cubed Numbers in Python

The challenge

Find the sum of the odd numbers within an array, after cubing the initial integers. The function should return None if any of the values aren’t numbers.

Note: Booleans should not be considered as numbers.

The solution in Python code

Option 1:

def cube_odd(arr): if any(type(x) is not int for x in arr): return None return sum(x ** 3 for x in arr if x % 2 != 0)
Code language: Python (python)

Option 2:

def cube_odd(arr): if len(set(map(type,arr))) < 2: return sum(n**3 for n in arr if n%2)
Code language: Python (python)

Option 3:

def cube_odd(arr): s = 0 for n in arr: if type(n) != int: break elif n%2: s += n**3 else: return s
Code language: Python (python)

Test cases to validate our solution

import test from solution import cube_odd @test.describe("Fixed Tests") def fixed_tests(): @test.it('Basic Test Cases') def basic_test_cases(): test.assert_equals(cube_odd([1, 2, 3, 4]), 28) test.assert_equals(cube_odd([-3,-2,2,3]), 0) test.assert_equals(cube_odd(["a",12,9,"z",42]), None) test.assert_equals(cube_odd([True,False,2,4,1]), None) @test.describe("Random tests") def _(): from random import randint, random sol=lambda arr: None if any(type(e)!=int for e in arr) else sum(e*e*e for e in arr if e%2) base=["a","b","c",True,False] for _ in range(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 range(randint(5,10))],key=lambda a: random()) expected = sol(arr) @test.it(f"Testing for cube_odd({arr})") def _(): test.assert_equals(cube_odd(arr[:]),expected)
Code language: Python (python)
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