Question: How to write a function that takes n (where n > 0) and returns the list of all combinations of positive integers that sum to n? This is a common question on the web. And there are different answers provided such as 1, 2 and 3. However, in the answers provided, they use two functions to solve the problem. I want to do it with only one single function. Therefore, I coded as follows: def all_combinations_sum_to_n(n): from itertools import combinations_with_replacement combinations_list = [] if n < 1: return combinations_list l = [i for i in range(1, n + 1)] for i in range(1, n + 1): combinations_list = combinations_list + (list(combinations_with_replacement(l, i))) result = [list(i) for i in combinations_list if sum(i) == n] result.sort() return result  If I pass 20 to my function which is all_combinations_sum_to_n(20), the OS of my machine kills the process as it is very costly. I think the space complexity of my function is O(n*n!). How do I modify my code so that I don't have to create any other function and yet my single function has an improved time or space complexity? I don't think it is possible by using itertools.combinations_with_replacement. UPDATE All answers provided by Barmar, ShadowRanger and pts are great. As I was looking for an efficient answer in terms of both memory and runtime, I used https://perfpy.com and selected python 3.8 to compare the answers. I used six different values of n and in all cases, ShadowRanger's solution had the highest score. Therefore, I chose ShadowRanger's answer as the best one. The scores were as follows: 

Answer: You've got two main problems, one causing your current problem (out of memory) and one that will continue the problem even if you solve that one:  You're accumulating all combinations before filtering, so your memory requirements are immense. You don't even need a single list if your function can be a generator (that is iterated to produce a value at a time) rather than returning a fully realized list, and even if you must return a list, you needn't generate such huge intermediate lists. You might think you need at least one list for sorting purposes, but combinations_with_replacement is already guaranteed to produce a predictable order based on the input ordering, and since range is ordered, the values produced will be ordered as well.  Even if you solve the memory problem, the computational cost of just generating that many combinations is prohibitive, due to poor scaling; for the memory, but not CPU, optimized version of the code below, it handles an input of 11 in 0.2 seconds, 12 in ~2.6 seconds, and 13 in ~11 seconds; at that scaling rate, 20 is going to approach heat death of the universe timeframes.   Barmar's answer is one solution to both problems, but it's still doing work eagerly and storing the complete results when the complete work might not even be needed, and it involves sorting and deduplication, which aren't strictly necessary if you're sufficiently careful about how you generate the results. This answer will fix both problems, first the memory issue, then the speed issue, without ever needing memory requirements above linear in n. Solving the memory issue alone actually makes for simpler code, that still uses your same basic strategy, but without consuming all your RAM needlessly. The trick is to write a generator function, that avoids storing more than one results at a time (the caller can listify if they know the output is small enough and they actually need it all at once, but typically, just looping over the generator is better): from collections import deque # Cheap way to just print the last few elements from itertools import combinations_with_replacement # Imports should be at top of file, # not repeated per call def all_combinations_sum_to_n(n): for i in range(1, n + 1): # For each possible length of combinations... # For each combination of that length... for comb in combinations_with_replacement(range(1, n + 1), i): if sum(comb) == n: # If the sum matches... yield list(comb) # yield the combination # 13 is the largest input that will complete in some vaguely reasonable time, ~10 seconds on TIO print(*deque(all_combinations_sum_to_n(13), maxlen=10), sep='\n')  Try it online! Again, to be clear, this will not complete in any reasonable amount of time for an input of 20; there's just too much redundant work being done, and the growth pattern for combinations scales with the factorial of the input; you must be more clever algorithmically. But for less intense problems, this pattern is simpler, faster, and dramatically more memory-efficient than a solution that builds up enormous lists and concatenates them. To solve in a reasonable period of time, using the same generator-based approach (but without itertools, which isn't practical here because you can't tell it to skip over combinations when you know they're useless), here's an adaptation of Barmar's answer that requires no sorting, produces no duplicates, and as a result, can produce the solution set in less than a 20th of a second, even for n = 20: def all_combinations_sum_to_n(n, *, max_seen=1): for i in range(max_seen, n // 2 + 1): for subtup in all_combinations_sum_to_n(n - i, max_seen=i): yield (i,) + subtup yield (n,) for x in all_combinations_sum_to_n(20): print(x)  Try it online! That not only produces the individual tuples with internally sorted order (1 is always before 2), but produces the sequence of tuples in sorted order (so looping over sorted(all_combinations_sum_to_n(20)) is equivalent to looping over all_combinations_sum_to_n(20) directly, the latter just avoids the temporary list and a no-op sorting pass).