Algorithm to find which number in a list sum up to a certain number

This problem reduces to the 0-1 Knapsack Problem, where you are trying to find a set with an exact sum. The solution depends on the constraints, in the general case this problem is NP-Complete.

However, if the maximum search sum (let’s call it S) is not too high, then you can solve the problem using dynamic programming. I will explain it using a recursive function and memoization, which is easier to understand than a bottom-up approach.

Let’s code a function f(v, i, S), such that it returns the number of subsets in v[i:] that sums exactly to S. To solve it recursively, first we have to analyze the base (i.e.: v[i:] is empty):

• S == 0: The only subset of [] has sum 0, so it is a valid subset. Because of this, the function should return 1.

• S != 0: As the only subset of [] has sum 0, there is not a valid subset. Because of this, the function should return 0.

Then, let’s analyze the recursive case (i.e.: v[i:] is not empty). There are two choices: include the number v[i] in the current subset, or not include it. If we include v[i], then we are looking subsets that have sum S - v[i], otherwise, we are still looking for subsets with sum S. The function f might be implemented in the following way:

def f(v, i, S):
  if i >= len(v): return 1 if S == 0 else 0
  count = f(v, i + 1, S)
  count += f(v, i + 1, S - v[i])
  return count

v = [1, 2, 3, 10]
sum = 12
print(f(v, 0, sum))

By checking f(v, 0, S) > 0, you can know if there is a solution to your problem. However, this code is too slow, each recursive call spawns two new calls, which leads to an O(2^n) algorithm. Now, we can apply memoization to make it run in time O(n*S), which is faster if S is not too big:

def f(v, i, S, memo):
  if i >= len(v): return 1 if S == 0 else 0
  if (i, S) not in memo:  # <-- Check if value has not been calculated.
    count = f(v, i + 1, S, memo)
    count += f(v, i + 1, S - v[i], memo)
    memo[(i, S)] = count  # <-- Memoize calculated result.
  return memo[(i, S)]     # <-- Return memoized value.

v = [1, 2, 3, 10]
sum = 12
memo = dict()
print(f(v, 0, sum, memo))

Now, it is possible to code a function g that returns one subset that sums S. To do this, it is enough to add elements only if there is at least one solution including them:

def f(v, i, S, memo):
  # ... same as before ...

def g(v, S, memo):
  subset = []
  for i, x in enumerate(v):
    # Check if there is still a solution if we include v[i]
    if f(v, i + 1, S - x, memo) > 0:
      subset.append(x)
      S -= x
  return subset

v = [1, 2, 3, 10]
sum = 12
memo = dict()
if f(v, 0, sum, memo) == 0: print("There are no valid subsets.")
else: print(g(v, sum, memo))

Disclaimer: This solution says there are two subsets of [10, 10] that sums 10. This is because it assumes that the first ten is different to the second ten. The algorithm can be fixed to assume that both tens are equal (and thus answer one), but that is a bit more complicated.