Dynamic programming aspect in Kadane's algorithm

Question

Initialize:
    max_so_far = 0
    max_ending_here = 0

Loop for each element of the array
   (a) max_ending_here = max_ending_here + a[i]
   (b) if(max_ending_here <         


        

Answer 1

Note that I derived my explanation from this answer. It demonstrates how Kadane’s algorithm can be seen as a DP algorithm which has overlapping subproblems.

Identifying subproblems and recurrence relations

Imagine we have an array a from which we want to get the maximum subarray. To determine the max subarray that ends at index i the following recursive relation holds:

max_subarray_to(i) = max(max_subarray_to(i - 1) + a[i], a[i])

In order to get the maximum subarray of a we need to compute max_subarray_to() for each index i in a and then take the max() from it:

max_subarray = max( for i=1 to n max_subarray_to(i) )

Example

Now, let's assume we have an array [10, -12, 11, 9] from which we want to get the maximum subarray. This would be the work required running Kadane's algorithm:

result = max(max_subarray_to(0), max_subarray_to(1), max_subarray_to(2), max_subarray_to(3))

max_subarray_to(0) = 10  # base case
max_subarray_to(1) = max(max_subarray_to(0) + (-12), -12)
max_subarray_to(2) = max(max_subarray_to(1) + 11, 11)
max_subarray_to(3) = max(max_subarray_to(2) + 9, 49)

As you can see, max_subarray_to() is evaluated twice for each i apart from the last index 3, thus showing that Kadane's algorithm does have overlapping subproblems.

Kadane's algorithm is usually implemented using a bottom up DP approach to take advantage of the overlapping subproblems and to only compute each subproblem once, hence turning it to O(n).