Maximum sum, min length subset

Question

I am trying to solve this popular problem:

"Given an integer array, divide the array into two subsets, A and B, while respecting the following conditions.

the intersection of A and B is null.
the union A and B is equal to the original array.
the number of elements in subset A is minimal.
the sum of A's elements is greater than the sum of B's elements.

Return the subset A in increasing order, where the sum of A's elements is greater than the sum of B's elements. If more than one subset exists, return the one with the maximal sum."

I found several solutions online but none of them solve it when the test case is [2,2,2,5]

I wrote the following code:

def subsetA(nums):
    nums.sort(reverse=True)
    subset_a = []
    sum_a = 0
    sum_b = 0

    for num in nums:
        if sum_a <= sum_b:
            sum_a += num
            subset_a.append(num)
        else:
            sum_b += num

    return sorted(subset_a)

This solution works for a lot of cases but it doesn't work for the test case I mentioned above.

The answer should be [2,2,2]. But i cannot think of any solution that will return that answer.

How can I improve this code?

also, we talk about "sublists" here, not "subsets", right? Subsets of [2,2,2,5] would be only {2}, {5} and {2,5}. — gog
– gog, Commented Dec 16, 2023 at 0:55

Cary Swoveland · Accepted Answer · 2023-12-16 00:22:07Z

1

One way to solve this is as a linear program with binary variables. Suppose the array is held by the variable arr, where arr_i is the value of the element at index i.

First define the binary variables.

x_i equals 1 if arr_i is assigned to A, and equals 0 if it is assigned to B

The objective (function) and constraints are as follows::

min ∑_ix_i

subject to:

∑_iarr_ix_i >= ∑_iarr_i(1-x_i) + t

t is a pre-defined tolerance to ensure that a strict inequality is achieved (linear programs cannot have strict inequalities).

Notice that 1-x_i equals 1 if arr_i is assigned to B.

The constraint needs to be rearranged as follows.

∑_iarr_ix_i >= (∑_iarr_i + t)/2

An integer linear programming (ILP) package can then be used to obtain optimal values for the variables.

edited Dec 16, 2023 at 0:22

answered Dec 16, 2023 at 0:14

Cary Swoveland

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Comments

MBo · Accepted Answer · 2023-12-16 04:24:23Z

0

You are using greedy algorithm, it does not guarantee the best solution. If sum of list is reliable, you can apply dynamic programming.

k-th entry of list a contains tuple (last index, number of items to compose sum k).

We fill table a: for every value n from nums check a entries whether a) sum k might be composed with n + k-n sum b) number of items for this sum is better than before

Then look for the smallest number of items in the second half of the table - it is the best sum for given condition. Then we retrieve used items.

def bestsubsetsum(nums):
    l = len(nums)
    S = sum(nums)
    nums.sort()
    a =[(0,l+1)]*(S+1)
    a[0] = (-1,0)

    for j, n in enumerate(nums):
        for k in range(S, n-1,-1):
            if a[k-n][0] and a[k-n][1]+1 < a[k][1]:
                a[k] = (j+1, a[k-n][1]+1)
    #print(a)

    a[0] = (-1, l+1)
    imin = 0
    for k in range(S//2+1,S+1):
        if a[k][1] <= a[imin][1]:
            imin = k
    A = []
    while imin:
        A.insert(0, nums[a[imin][0]-1])
        imin -= nums[a[imin][0]-1]
    return A


print(bestsubsetsum([3,7,12,1]))
print(bestsubsetsum([2,2,2,5]))
print(bestsubsetsum([2,4,3,7,3,7,3]))
print(bestsubsetsum([2,4,3,9,3,7,3]))

>>
[12]
[2, 5]
[4, 7, 7]
[7, 9]

edited Dec 16, 2023 at 4:24

answered Dec 16, 2023 at 3:33

MBo

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5 Comments

Jonathan Over a year ago

This solution doesn't work. print(bestsubsetsum([1,1,1,2])) returns [1,2]. This is wrong, it should retain [1,1,1,2]

MBo Over a year ago

No, this is correct solution. 1,2 is the shortest subset with sum larger than rest

Jonathan Over a year ago

For input [2,2,2,5], subset a contains [2,5] and b contains [2, 2]. This breaks condition #1. The correct answer should be a = [2,2,2] and b= [5].

MBo Over a year ago

It seems that "set" term is used incorrectly , because we cannot build source array from true sets in many cases

Kelly Bundy Over a year ago

@Jonathan Why should it return [1,1,1,2], why not [1,1,1]?

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