Find sum in array equal to zero - algorithm

Question

Given an array of integers, find a set of at least one integer which sums to 0.
For example, given [-1, 8, 6, 7, 2, 1, -2, -5], the algorithm may output [-1, 6, 2, -2, -5] because this is a subset of the input array, which sums to 0.
The solution must run in polynomial time.

Answer 1

You'll have a hard time doing this in polynomial time, as the problem is known as the Subset sum problem, and is known to be NP-complete.
If you do find a polynomial solution, though, you'll have solved the "P = NP?" problem, which will make you quite rich.
The closest you get to a known polynomial solution is an approximation, such as the one listed on Wikipedia, which will try to get you an answer with a sum close to, but not necessarily equal to, 0.

Answer 2

This is a Subset sum problem, It's NP-Compelete but there is pseudo polynomial time algorithm for it. see wiki.
The problem can be solved in polynomial if the sum of items in set is polynomially related to number of items, from wiki:
The problem can be solved as follows
using dynamic programming. Suppose the
sequence is
x1, ..., xn
and we wish to determine if there is a
nonempty subset which sums to 0. Let N
be the sum of the negative values and
P the sum of the positive values.
Define the boolean-valued function
Q(i,s) to be the value (true or false)
of
"there is a nonempty subset of x1, ..., xi which sums to s".
Thus, the solution to the problem is
the value of Q(n,0).
Clearly, Q(i,s) = false if s < N or s
P so these values do not need to be stored or computed. Create an array to
hold the values Q(i,s) for 1 ≤ i ≤ n
and N ≤ s ≤ P.
The array can now be filled in using a
simple recursion. Initially, for N ≤ s
≤ P, set
Q(1,s) := (x1 = s).
Then, for i = 2, …, n, set
Q(i,s) := Q(i − 1,s) or (xi = s) or Q(i − 1,s − xi) for N ≤ s ≤ P.
For each assignment, the values of Q
on the right side are already known,
either because they were stored in the
table for the previous value of i or
because Q(i − 1,s − xi) = false if s −
xi < N or s − xi > P. Therefore, the
total number of arithmetic operations
is O(n(P − N)). For example, if all
the values are O(nk) for some k, then
the time required is O(nk+2).
This algorithm is easily modified to
return the subset with sum 0 if there
is one.
This solution does not count as
polynomial time in complexity theory
because P − N is not polynomial in the
size of the problem, which is the
number of bits used to represent it.
This algorithm is polynomial in the
values of N and P, which are
exponential in their numbers of bits.
A more general problem asks for a
subset summing to a specified value
(not necessarily 0). It can be solved
by a simple modification of the
algorithm above. For the case that
each xi is positive and bounded by the
same constant, Pisinger found a linear
time algorithm.[2]

Answer 3

It is well known Subset sum problem which NP-complete problem.
If you are interested in algorithms then most probably you are math enthusiast that I advise you look at
Subset Sum problem in mathworld
and here you can find the algorithm for it
Polynomial time approximation algorithm
initialize a list S to contain one element 0.
for each i from 1 to N do
let T be a list consisting of xi+y,
for all y in S
let U be the union of T and S
sort U
make S empty
let y be the smallest element of U
add y to S
for each element z of U in
increasing order do //trim the list by
eliminating numbers
close one to another
if y<(1-c/N)z, set y=z and add z to S
if S contains a number between (1-c)s and s, output yes, otherwise no