First clear solution for Index Power by bukebuer - python coding challenges

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First solution in Clear category for Index Power by bukebuer

def index_power(array, n):
    return array[n]**n if len(array)>n else -1

Sept. 3, 2014

Comments:

ShoddyFirstTimer on Oct. 13, 2014, 7:49 p.m.
My code is similar to this but on more than one line, lol, this is very interesting.

bart.fiten on Nov. 10, 2014, 7:56 p.m.
Wouldn't this give an error if n is on the last index in the array? array = [1,2,3,4,5] -> len(array) = 5 So if you would take 5 for n, there is nothing on this index location, right? Shouldn't the solution be this: return array[n]**n if (len(array) - 1) > n else -1

bukebuer on Nov. 11, 2014, 12:57 a.m.
When len(array)=5, then maximum feasible n can only be 4, because index of list begins from 0. n=5 -> len(array)=5>5 is False -> return -1

gotoloop on Nov. 15, 2014, 8:13 p.m.
Funny... mine was the same but returning -1 1st! :D <h2>return -1 if n >= len(array) else array[n]**n</h2>

sct10 on Feb. 27, 2015, 9:53 p.m.
Error here. If n is just equal to the length you will get an out of range error.,

lennon.mary.e on Nov. 2, 2016, 5:48 p.m.
Agreed, it needs: def index_power(array, n): return array[n]**n if len(array) >= n else -1

serseregas777 on April 25, 2015, 4:10 a.m.
This is very interesting way to solve this problem!

Nokfi on Aug. 15, 2015, 9:13 p.m.
It is not given that N is an int, neither is it given that N >=0. This solution does not check this.

kirya05 on Dec. 30, 2019, 6:43 p.m.
follow the conditions for writing operands

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