带有替代符号的不相交的maxsubarray

Question

i would like to find an algorithm to find the max disjoint subarray with alternated signs (not necessarily contiguos), for example in (2, -3, 4., 5, -4, 3, -3, 5, -2, 1) it returns 7 that is the greatest sum, and the subarray is (5,-3,5)

I tried something like this using dp:

    A=[2,-3,4,5,-4,3,-3,5,-2,1]

    m = A[0]
    flag = A[0]    #flag has the same sign of the previously taken element
    maximum = m    #temporary maxsum

    for i in range(1,10):
        if A[i]>0 and flag<0 or A[i]<0 and flag>0: #i look only for alternate signs
             m = max(m ,A[i]+m)
             if m > maximum:
                   flag = -flag
                    maximum = m
             else:
                   if A[i]>maximum:
                   maximum=A[i]
                   flag=-flag


     print(maximum)

it gives 7 but it's only a coincidence

Should i have to use another For nested in order to properly compare every possible subarray?

Answer 1

A=[2,-3,4,5,-4,3,-3,5,-2,1]
dp = [[([], 0), ([], 0)]]
for i, x in enumerate(A):
    curr = list(dp[-1])
    if x < 0 and curr[0][1] + x > curr[1][1]:
        curr[1] = (curr[0][0] + [i], curr[0][1] + x)
    elif x > 0 and curr[1][1] + x > curr[0][1]:
        curr[0] = (curr[1][0] + [i], curr[1][1] + x)
    dp.append(curr)

print dp[-1][0]

This will print a tuple with the indices of the elements to take, and the maximum sum.

Each element in dp represents the best solution up to that point in the list for the case where the next number needs to be positive (first element) and where the next number needs to be negative (second element). Within each of those best solutions, we a list of indices to get to that point, and the maximum sum. The answer is always going to be from the last element in dp where the most recently taken number was positive.

If you don't need to track the elements used:

A=[2,-3,4,5,-4,3,-3,5,-2,1]
dp = [[0, 0]]
for i, x in enumerate(A):
    curr = list(dp[-1])
    if x < 0 and curr[0] + x > curr[1]:
        curr[1] = curr[0] + x
    elif x > 0 and curr[1] + x > curr[0]:
        curr[0] = curr[1] + x
    dp.append(curr)

print dp[-1][0]