How to partition an array of integers in a way that minimizes the maximum of the sum of each partition?

Question

The inputs are an array A of positive or null integers and another integer K.

We should partition A into K blocks of consecutive elements (by "partition" I mean that every element of A belongs to some block and 2 different blocks don't contain any element in common).

We define the sum of a block as sum of the elements of the block.

The goal is to find such a partition in K blocks such that the maximum of the sums of each block (let's call that " MaxSumBlock ") is minimized.

We need to output the MaxSumBlock (we don't need to find an actual partition)

Here is an example:

Input:

A = {2, 1, 5, 1, 2, 2, 2}
K = 3

Expected output:

MaxSumBlock: 6
(with partition: {2, 1}, {5, 1}, {2, 2, 2})

In the expected output, the sums of each block are 3, 6 and 6. The max is 6.

Here is an non optimal partition:

partition: {2, 1}, {5}, {1, 2, 2, 2}

The sums of each block in that case are 3, 6 and 7. The max is hence 7. It is not a correct answer.

What algorithm solves this problem?

EDIT: K and the size of A is no bigger than 100'000. Each element of A is no bigger than 10'000

Answer 1

Use binary search.

Let max sum range from 0 to sum(array). So, mid = (range / 2). See if mid can be achieved by partitioning into k sets in O(n) time. If yes, go for lower range and if not, go for a higher range.

This will give you the result in O(n log n).

PS: if you have any problem with writing the code, I can help but I'd suggest you try it first yourself.

EDIT:
as requested, I'll explain how to find if mid can be achieved by partitioning into k sets in O(n) time.
Iterate through the elements till sum is less than or equal to mid . As soon as it gets greater than mid , let it be part of next set. If you get k or less sets, mid is achievable, else not.