Dividing an array of n integers into M continuous sub sequences so that the maximal sum of some partition is minimal

Question

First of all I must mention I've seen some solutions on this site that doesn't work after I checked them or do not match my question since I want to do it in a dynamic programming way.

I've tried all things but I'm really lost, I need to partition the given sequence of integers into M continous sub sequences so that the maximal sum of some partition is maximal.

For example: {4,8,7,10,40,15,30,2,1,20} and M = 4 The algorithm should generate {4,8,7,10}{40}{15,30}{2,1,20} because now the maximum sum is 15 + 30 = 45. Again, I mention that I've seen a post on this site which didn't work. Thank you for all the help!

Answer 1

A straight forward dynamic programming solutions:

1. The state is (i, j) , where i is the last element of the current partition and j is the number of parts there're in partition.

2. The value of the state is the maximum sum so far.

3. The transitions are max(f(i, j), sum(i + 1, k)) -> f(k, j + 1) (it means that we take the [i + 1, k] part).

4. The answer is f(N, M) (it means that we took N elements and created exactly M groups).

Answer 2

The maximum sum partition can be found by taking the entire sequence and cutting off M-1 one-element partitions on either side. You should do this in a way such that the sum of cut-off elements is minimal.

This gives you a single integer degree of freedom (the number of elements to cut off from the beginning). So you can just iterate this number from 0 thru M - 1 (the number of elements to cut off from the end is then obviously M - 1 - i ) and take the minimum.

In your example, this would result in:

{4}, {8}, {7}, {10, 40, 15, 30, 2, 1, 20}

The maximum sum of any partition is then 118.