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First solution in Clear category for Index Power by maxnrg
"""Index Power
You are given an array with positive numbers and a number N.
You should find the N-th power of the element in the array with the index N.
If N is outside of the array, then return -1.
Don't forget that the first element has the index 0.
Let's look at a few examples:
- array = [1, 2, 3, 4] and N = 2, then the result is 32 == 9;
- array = [1, 2, 3] and N = 3, but N is outside of the array, so the result is -1.
Input: Two arguments. An array as a list of integers and a number as a integer.
Output: The result as an integer.
Example:
index_power([1, 2, 3, 4], 2) == 9
index_power([1, 3, 10, 100], 3) == 1000000
index_power([0, 1], 0) == 1
index_power([1, 2], 3) == -1
"""
def index_power(array: list,
n: int) -> int:
""" Find Nth power of the element with index N. """
if n >= len(array):
return -1
return array[n] ** n
if __name__ == '__main__':
print('Example:')
print(index_power([1, 2, 3, 4], 2))
# 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"
print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")
Dec. 14, 2019
