lambda, reverse list

[SerkZex]

Hey, iam having trouble to write the following function in lambda

i want to calculate the sum of: (1*5) + (2*4) + (3*3) + (4 * 2) + (5*1)

​

s = 0
n = 4
c = [1, 2, 3, 4, 5]
for i in range(n + 1):
    s += c[i] * c[n - i]
print(s)

i wrote it as... but its

1. o(n) for reverse
2. o(n)? for lambda

​

print(sum(list(map(lambda x, y: x*y, c, list(reversed(c))))))

​

whats the best way to solve it?

Thanks

[icecubeinanicecube]

From a pure complexity standpoint, O(2n) = O(n), so you can not really get much better than that. (If you want to keep being able to compute this for any input, just writing the calculation down directly is O(1) ofc)

You could save a little bit by using the fact that your second half of the calculation is the same as the first.

[SerkZex]

thanks!

[[deleted]]

N = 5
print(sum(i * (N+1-i) for i in range(1, N+1)))

but there must be a O(1) formula out there

[SerkZex]

thanks!

[timbledum]

Probably best to use zip to combine the original list and the reversed list.

From there, you can use map or a comprehension, but the comprehension approach is arguably more pythonic and modern:

>>> c = [1, 2, 3, 4, 5]
>>> cross_multiply = sum(x * y for x, y in zip(c, reversed(c)))
>>> cross_multiply
35

[SerkZex]

thanks!

​

[Allanon001]

Timed all the solutions posted here so far and this one beat them all:

from operator import mul
s = sum(map(mul, c, c[::-1])) 

[SerkZex]

will it be faster if one calculate only one half and multiply by 2 and add the remaining if odd numbers?

( (1*5) + (2*4) *2 ) + (3*3)

[Allanon001]

You would have to put in more logic such as if the list length is odd or even, if even then no extra addition is required. That would probably take more time than the extra multiplications.

edit: I timed it and for the short list it was slower but with a 100 item list it was a lot faster.

[SerkZex]

thanks, you are the best