Maxumum non-contigious subarray in Python

Question

I have to fing an increasing non-contigious subarray with maximum sum using Python. For example, if I have an array [5, 2, 4, 3, 7, 7], than I have to output [2, 4, 7]. Output given is increasing, have the maximum sum among all the possible increasing subarrays.

@CarloZanocco A non decreasing subsequence with maximum sum, although the question is poorly written. — nice_dev, Sep 29 '20 at 12:58
@fiendfire2807 can you tell me if I have correctly understood your question? — Carlo Zanocco, Sep 29 '20 at 13:17
Does this answer your question? [Maximum Sum of a non decreasing sub-sequence in an array using fenwick tree or BIT](https://stackoverflow.com/questions/15193469/maximum-sum-of-a-non-decreasing-sub-sequence-in-an-array-using-fenwick-tree-or-b) — Stef, Sep 29 '20 at 14:20
The accepted answer has complexity O(n^2) both in time and space, but this question is actually a duplicate and the previous question has an answer is O(n log n) time and O(n) space: [maximum sum of a non-decreasing sub-sequence in an array](https://stackoverflow.com/questions/15193469/maximum-sum-of-a-non-decreasing-sub-sequence-in-an-array-using-fenwick-tree-or-b) — Stef, Sep 29 '20 at 14:21
@Stef The question that you consider duplicate require another type of output, here we need a list not the max value as int — Carlo Zanocco, Sep 29 '20 at 15:48
@CarloZanocco Those two problems are pretty close; deriving the "path" to the optimal solution found with a dynamic programming algorithm is usually very easy, by tracing the optimal values in the array used by the algorithm. — Stef, Sep 29 '20 at 16:19
@Stef Have you tried to run the code? `[5, 2, 4, 3, 7, 7]` return `20` insed of `13` — Carlo Zanocco, Sep 29 '20 at 16:58

Carlo Zanocco · Accepted Answer · 2020-09-29T13:18:19.287

Take a look at this code:

def elaborate(A):# Iterative function to princreasing subsequence with the maximum sum
    n = len(A)

    # values[i] stores the increasing subsequence having maximum sum
    # that ends with A[i]
    values = [[] for _ in range(n)]
    values[0].append(A[0])

    # sum[i] stores the maximum sum of the increasing subsequence
    # that ends with A[i]
    sum = [0] * n
    sum[0] = A[0]

    for i in range(1, n): # start from second element in the list
        for j in range(i):# do for each element in sublist[0..i-1]
            # find increasing subsequence with maximum sum that ends with
            # A[j] where A[j] is less than the current element A[i]
            if sum[i] < sum[j] and A[i] > A[j]:
                values[i] = values[j].copy()    # update increasing subsequence
                sum[i] = sum[j]     # update maximum sum

        values[i].append(A[i])  # include current element in increasing subsequence
        sum[i] += A[i]  # add current element to maximum sum

    j = 0    # j will contain index of values
    for i in range(1, n):
        if sum[i] > sum[j]:
            j = i

    # print values
    print(values[j])

A = [5, 2, 4, 3, 7, 7] 
elaborate(A)

# OUTPUT: [2, 4, 7]

The time complexity of above solution is O(n^2) and auxiliary space used by the program is O(n^2)

Maxumum non-contigious subarray in Python

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