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

Question

I have a list of numbers. I also have a certain sum. The sum is made from a few numbers from my list (I may/may not know how many numbers it's made from). Is there a fast algorithm to get a list of possible numbers? Written in Python would be great, but pseudo-code's good too. (I can't yet read anything other than Python :P )

Example

list = [1,2,3,10]
sum = 12
result = [2,10]

NOTE: I do know of Algorithm to find which numbers from a list of size n sum to another number (but I cannot read C# and I'm unable to check if it works for my needs. I'm on Linux and I tried using Mono but I get errors and I can't figure out how to work C# :(
AND I do know of algorithm to sum up a list of numbers for all combinations (but it seems to be fairly inefficient. I don't need all combinations.)

Spanos

Googling for "subset sum" will probably give some useful results. – Hart

As a side note, if you know Python well it shouldn't be that hard to read languages like C# and at least figure out the big picture of what the code is doing. – Filet

Concerning > I don't need all combinations: Since this problem is known to be NP-complete, at last you probably will have to enumerate all possibilities. – Numbing

@musicfreak: I'm still in the learning stages. I attempted to rewrite it in Python but it didn't seem to work with a set of 4 numbers and 1 sum; so I'm assuming I didn't write it right. – Spanos

Answer 1

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.

Councilman

Thanks! That's exactly what I was looking for. I've never done stuff this advanced so this is great! – Spanos

You're welcome =). If you like Dynamic Programming, there is a nice tutorial at topcoder.com/tc?module=Static&d1=tutorials&d2=dynProg. – Councilman

I'm trying to translate your code to ruby, but I am not having much luck at the moment. Here is my attempt: gist.github.com/webconsult/8710eede3f91d84d7860 Can anyone help me figure out what I am doing wrong? It reports undefined method `+' for nil:NilClass (on line 5), but debugging reveals that it only happens once the recursive call on line 6 is triggered. I'm a bit confused as to what is going on? – Stratagem

I tried this with a 1M long list. And I ran into an maximum recursion depth exceeded RuntimeError – Dejection

Hi does anyone know how to get ALL the different solution that adds up to the same total... using the slightly modified code of above.... for example: with v = [1100, 1105, 11830, 14790, 2325, 2455, 2555, 2935, 3050, 3150, 3185, 3370, 3475, 350, 3530, 3590, 3680, 3745, 885, 9624] sum = 43029... there are multiple solutions.. and i want to get all the solutions.. please advise. – Xenophobe

Thanks for saving me some time, although the use of a keyword as a variable (sum) can result in problems. – Hardy

@Councilman This code is checking for sum== 10 how i can check for sum>=10 ? Kindly help – Barbed

Answer 2

I know I'm giving an answer 10 years later since you asked this, but i really needed to know how to do this an the way jbernadas did it was too hard for me, so i googled it for an hour and I found a python library itertools that gets the job done!

I hope this help to future newbie programmers. You just have to import the library and use the .combinations() method, it is that simple, it returns all the subsets in a set with order, I mean:

For the set [1, 2, 3, 4] and a subset with length 3 it will not return [1, 2, 3][1, 3, 2][2, 3, 1] it will return just [1, 2, 3]

As you want ALL the subsets of a set you can iterate it:

import itertools

sequence = [1, 2, 3, 4]
for i in range(len(sequence)):
    for j in itertools.combinations(sequence, i):
        print(j)

The output will be

() (1,) (2,) (3,) (4,) (1, 2) (1, 3) (1, 4) (2, 3) (2, 4) (3, 4) (1, 2, 3) (1, 2, 4) (1, 3, 4) (2, 3, 4)

Hope this help!

Dingdong

Thanks for answering. Literally saved me a lot of time :) – Jahdol

It doesn't work for sequence [1, 2] and target sum 3 – Sachikosachs

from big(O) standpoint this is o(n2). I am wondering if there is an efficient solution – Cenotaph

Answer 3

So, the logic is to reverse sort the numbers,and suppose the list of numbers is l and sum to be formed is s.

   for i in b:
            if(a(round(n-i,2),b[b.index(i)+1:])):
                r.append(i)    
                return True
        return False

then, we go through this loop and a number is selected from l in order and let say it is i . there are 2 possible cases either i is the part of sum or not. So, we assume that i is part of solution and then the problem reduces to l being l[l.index(i+1):] and s being s-i so, if our function is a(l,s) then we call a(l[l.index(i+1):] ,s-i). and if i is not a part of s then we have to form s from l[l.index(i+1):] list. So it is similar in both the cases , only change is if i is part of s, then s=s-i and otherwise s=s only.

now to reduce the problem such that in case numbers in l are greater than s we remove them to reduce the complexity until l is empty and in that case the numbers which are selected are not a part of our solution and we return false.

if(len(b)==0):
    return False    
while(b[0]>n):
    b.remove(b[0])
    if(len(b)==0):
        return False    

and in case l has only 1 element left then either it can be part of s then we return true or it is not then we return false and loop will go through other number.

if(b[0]==n):
    r.append(b[0])
    return True
if(len(b)==1):
    return False

note in the loop if have used b..but b is our list only.and i have rounded wherever it is possible, so that we should not get wrong answer due to floating point calculations in python.

r=[]
list_of_numbers=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]
list_of_numbers=sorted(list_of_numbers)
list_of_numbers.reverse()
sum_to_be_formed=401.54
def a(n,b):
    global r
    if(len(b)==0):
        return False    
    while(b[0]>n):
        b.remove(b[0])
        if(len(b)==0):
            return False    
    if(b[0]==n):
        r.append(b[0])
        return True
    if(len(b)==1):
        return False
    for i in b:
        if(a(round(n-i,2),b[b.index(i)+1:])):
            r.append(i)    
            return True
    return False
if(a(sum_to_be_formed,list_of_numbers)):
    print(r)

this solution works fast.more fast than one explained above. However this works for positive numbers only. However also it works good if there is a solution only otherwise it takes to much time to get out of loops.

an example run is like this lets say

    l=[1,6,7,8,10]

and s=22 i.e. s=1+6+7+8
so it goes through like this 

1.) [10, 8, 7, 6, 1] 22
i.e. 10  is selected to be part of 22..so s=22-10=12 and l=l.remove(10)
2.) [8, 7, 6, 1] 12
i.e. 8  is selected to be part of 12..so s=12-8=4 and l=l.remove(8)
3.) [7, 6, 1] 4  
now 7,6 are removed and 1!=4 so it will return false for this execution where 8 is selected.
4.)[6, 1] 5
i.e. 7  is selected to be part of 12..so s=12-7=5 and l=l.remove(7)
now 6 are removed and 1!=5 so it will return false for this execution where 7 is selected.
5.)[1] 6
i.e. 6  is selected to be part of 12..so s=12-6=6 and l=l.remove(6)
now 1!=6 so it will return false for this execution where 6 is selected.
6.)[] 11
i.e. 1 is selected to be part of 12..so s=12-1=1 and l=l.remove(1)
now l is empty so all the cases for which 10 was a part of s are false and so 10 is not a part of s and we now start with 8 and same cases follow.
7.)[7, 6, 1] 14
8.)[6, 1] 7
9.)[1] 1

just to give a comparison which i ran on my computer which is not so good. using

l=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,145.21,123.56,11.90,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]

and

s=2000

my loop ran 1018 times and 31 ms.

and previous code loop ran 3415587 times and took somewhere near 16 seconds.

however in case a solution does not exist my code ran more than few minutes so i stopped it and previous code ran near around 17 ms only and previous code works with negative numbers also.

so i thing some improvements can be done.

Geometer

While this code may work well, a good answer would include an explanation of how it works and why it is a good solution. – Looney

Answer 4

#!/usr/bin/python2

ylist = [1, 2, 3, 4, 5, 6, 7, 9, 2, 5, 3, -1]
print ylist 
target = int(raw_input("enter the target number")) 
for i in xrange(len(ylist)):
    sno = target-ylist[i]
    for j in xrange(i+1, len(ylist)):
        if ylist[j] == sno:
            print ylist[i], ylist[j]

This python code do what you asked, it will print the unique pair of numbers whose sum is equal to the target variable.

if target number is 8, it will print: 
1 7
2 6
3 5
3 5
5 3
6 2
9 -1
5 3

Roundtheclock

This is great. It does exit silently if no results are found. – Rural

What if the sum you're looking for is 22? – Hendrix

Answer 5

I have found an answer which has run-time complexity O(n) and space complexity about O(2n), where n is the length of the list.

The answer satisfies the following constraints:

1. List can contain duplicates, e.g. [1,1,1,2,3] and you want to find pairs sum to 2

2. List can contain both positive and negative integers

The code is as below, and followed by the explanation:

def countPairs(k, a):
    # List a, sum is k
    temp = dict()
    count = 0
    for iter1 in a:
        temp[iter1] = 0
        temp[k-iter1] = 0
    for iter2 in a:
        temp[iter2] += 1
    for iter3 in list(temp.keys()):
        if iter3 == k / 2 and temp[iter3] > 1:
            count += temp[iter3] * (temp[k-iter3] - 1) / 2
        elif iter3 == k / 2 and temp[iter3] <= 1:
            continue
        else:
            count += temp[iter3] * temp[k-iter3] / 2
    return int(count)

1. Create an empty dictionary, iterate through the list and put all the possible keys in the dict with initial value 0. Note that the key (k-iter1) is necessary to specify, e.g. if the list contains 1 but not contains 4, and the sum is 5. Then when we look at 1, we would like to find how many 4 do we have, but if 4 is not in the dict, then it will raise an error.
2. Iterate through the list again, and count how many times that each integer occurs and store the results to the dict.

3. Iterate through through the dict, this time is to find how many pairs do we have. We need to consider 3 conditions:

   3.1 The key is just half of the sum and this key occurs more than once in the list, e.g. list is [1,1,1], sum is 2. We treat this special condition as what the code does.

   3.2 The key is just half of the sum and this key occurs only once in the list, we skip this condition.

   3.3 For other cases that key is not half of the sum, just multiply the its value with another key's value where these two keys sum to the given value. E.g. If sum is 6, we multiply temp[1] and temp[5], temp[2] and temp[4], etc... (I didn't list cases where numbers are negative, but idea is the same.)

The most complex step is step 3, which involves searching the dictionary, but as searching the dictionary is usually fast, nearly constant complexity. (Although worst case is O(n), but should not happen for integer keys.) Thus, with assuming the searching is constant complexity, the total complexity is O(n) as we only iterate the list many times separately.

Advice for a better solution is welcomed :)

Grogshop