Need an algorithm for this problem

Question

There are two integer sequences A[] and B[] of length N,both unsorted.

Requirement: through the swapping of elements between A[] and B[]( can randomly exchange, not with same index), make the difference between {the sum of all elements in A[]} and {the sum of all elements in B[]} to be minimum.

PS: actually,it is an interview question I encountered.

Many thanks

Answer 1

"sequences A[] and B[] of length N" -> does this mean both A and B are each of length N?

(For the purpose of clarity I am using 1-based arrays below).

If so, how about this:

1. Assume A[1..N] and B[1..N]
2. Concatenate A and B into a new array C of length 2N: C[1..N] <- A[1..N]; C[N+1 .. 2N] <- B[1..N]
3. Sort C in ascending order.
4. Take the first pair of numbers from C; send the first element (C[1]) to A[1] and second element (C[2]) to B[1]
5. Take the second pair of numbers from C; this time send the second element (C[4]) to A[2] and the first element (C[3]) to B[2] (the order of elements in the pair sent to A and B is the opposite of 3)
6. ... repeat 3 and 4 until C is exhausted

The observation here is that, in a sorted array, an adjacent pair of numbers will have the smallest difference (compared to a pair of numbers from non-adjacent positions). Step 3 ensures that A[1] and B[1] consists of a pair of numbers with the least possible difference. Step 4 ensures that (a) A[2] and B[2] consist of a pair of numbers with the least possible difference (from the available numbers) and also (b) that the difference is opposite in sign from step 3. By continuing like this, we are ensuring that A[i] and B[i] contain numbers with the least possible difference. Also, by flipping the order in which we send elements to A and B, we are ensuring that the difference changes sign for each successive i.

Answer 2

The problem is NP-Complete.

We can reduce the partition problem to the decision version of this problem, i.e. given two arrays of ints of the same size, determine whether items can be swapped so that the sums are equal.

The input to the partition problem: a set S of integers, of size N

In order to transform this input into an input to our problem, we define A to be an array of all items in S, and B an array of the same size, with B[i]=0 for all i. This transformation is linear in the input size.

It is clear that our algorithm applied on A and B returns true if and only if there is a partition of S into 2 subsets such that the sums are equal.

Answer 3

This is going to be NP-hard! I believe you can do a reduction from Subset Sum to this.

As per BlueRaja/polygene's comments, I will try to provide a full reduction from Subset Sum.

Here is a reduction:

Subset Sum problem: Given integers x₁, x₂, ..., x_n, is there some non-empty subset which sums to zero?

Our problem: Given two integer arrays of size k, find the minimum possible difference of the sum of the two arrays, assuming we can shuffle around the integers in the arrays, treating both arrays as one array.

Say we had a polynomial time algo for our problem.

Say now you are given integers T = {x₁,x₂, ...,x_n} (multiset)

Let S_i = x₁ + x₂ + ...+ x_n + x_i.

Let T_i = {x₁, x₂, ..., x_i-1, x_i+1, ..., x_n } ( = T - x_i)

Define

A_i = Array formed using T_i

B_i = [S_i, 0, ..., 0] (i.e one element is S_i and rest are zeroes).

Let m_i = the min difference found by our problem for arrays A_i and B_i

(we run our problem n times).

Claim: Some non-empty subset of T sums to zero if and only if, there is some i, for which m_i = 0.

Proof: (wlog) say x₁ + x₂ + .. + x_k = 0

Then

A = [x_k+1, ..., x_n, 0, ...0]

B = [x₂, x₃, ..., x_k, S₁, 0, ..0]

gives the minimum difference m₁ to be |x₂ + .. + x_k + (x₁ + ... + x_n) + x₁ - (x_k+1 + .. + x_n)| = |2(x₁+ x₂ + .. x_k)| = 0.

Similarly the if part can be proved.

In fact, this actually also follows (more easily) from Partition too: just create new array with all zeroes.

Hoepfully I haven't made any mistakes.

Answer 4

Take any instance of the NP-complete partition problem:

Partition a multiset A of positive integers into two multisets B and C with the same sum

like {a₁,a₂,...,a_n}. Add n zeroes {0,0,0...,0,a₁,...,a_n} and ask if the set can be partitioned into two multisets A and B with the same sum and same number of elements. I claim these two conditions are equivalent:

• If A and B are a solution to the problem, then you can strike out the zeroes and get a solution of partiton problem.
• If there is a solution to the partition problem, for example a_i1 + a_i2 + ... a_ik = a_j1 + ... +a_jl where {a_i1, a_i2, a_ik, a_j1, ..., a_jl} = {a₁, ... , a_n} then obviously k+l = n. Add l zeroes to the left side and k zeroes to the right side and you'll get 0 + ... + 0 + a_i1 + a_i2 + ... a_ik = 0 + ... + 0 + a_j1 + ... +a_jl, whichi is a solution of your problem.

So, this is a reduction (so the problem is NP-hard) and the problem is NP, so it is NP-complete.

Answer 5

I'm not sure that this will ensure the minimum possible distance, but the first thing that comes to mi mind is something like this:

int diff=0;
    for (int i = 0; i<len; i++){
        int x = a[i] - b[i];
        if (abs(diff - x) > abs(diff + x)){
             swap(a,b,i);
             diff-=x;
        }else{
             diff+=x;
        }

    }

assuming that you have a swap function which takes the two arrays and exchanges the items at position i :)

computing and adding the difference between the two values at position i you get the incremental difference between the sums of the elements of the two arrays.
at each step you check if it's better to add (a[i]-b[i]) or (b[i]-a[i]). if the b[i]-a[i] it's the case, you swap the elements at position i in the arrays.

Maybe this will not be the best way, but it should be a start :)

Answer 6

Try being greedy about it. Given such limited information, I'm not sure what else one could put out there.