How to reduce the memory consumed by itertools.product?

Question

I am trying to generate all the permutations of a range of numbers(a,b) of size k and count the number of permutations whose sum is even. Currently my code is

highLow = [int(i) for i in input().split()]
k = int(input())
permutations = list(product(list(range(highLow[0],highLow[1]+1)), repeat=k))
print(len(list(filter(lambda x:sum(x)%2 == 0, permutations))))

But this is giving me a memory limit exceeded. How to solve this memory problem?

Answer 1

product uses O(1) memory; it's only your use of list to create a list of all the values that uses so much memory.

low, high = [int(i) for i in input().split()]
k = int(input())
permutations = product(range(low, high + 1), repeat=k)
n = sum(1 for _ in filter(lambda x: sum(x) % 2 == 0, permutations))
print(n)

Answer 2

As chepner answered, product uses only O(1) memory. You can also solve your problem with O( 0 ) memory for product , though, by not using product at all.

For example, the answer for low, high = 1, 99 and k = 100 is 18301617063661475246530801328625869309485603831946184570297868634965852237536237409359827175501347520033078455032642163735911784840089970792855267724585378713694517503049135418557489109958380424745001. With that method of filtering/counting product , that would take an eternity. With a little math, I computed that in less than a millisecond.

Yeah, this doesn't exactly answer the question as asked, but addresses the actual task. It's so common that people try to go through all permutations/combinations when they really shouldn't, that I think this is a useful answer.