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First solution in Clear category for Index Power by Krzysztof__wierblewski
def index_power(array, n):
if n > len(array)-1:
x=-1
else:
x=array[n]**n
"""
Find Nth power of the element with index N.
"""
return x
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert index_power([1, 2, 3, 4], 2) == 9, "Square"
assert index_power([1, 3, 10, 100], 3) == 1000000, "Cube"
assert index_power([0, 1], 0) == 1, "Zero power"
assert index_power([1, 2], 3) == -1, "IndexError"
Nov. 6, 2016
