Our last solution had a cubic time complexity: `O(N^3)`

.

One iteration for the length of the list `O(N)`

, inside **another** iteration for the length of the list `O(N^2)`

, inside the inner loop, we copied a sub-list: `O(N^3)`

.

Let’s optimize using an important concept: **We don’t always need to create every possible outcome to know what’s best.**

Do we need to make every sub-list? No! We can visit each value within the list and keep a running tally of sums.

Let’s see what information we can gather in a single pass:

# <> will mark the current element nums = [1, -7, 2, 15, -11, 2] most_seen = 0 current_max_sub_sum = 0 [<1>, -7, 2, 15, -11, 2] most_seen = 1 current_max_sub_sum = 1 [1, <-7>, 2, 15, -11, 2] most_seen = 1 current_max_sub_sum = -6 # our sum is negative # anything positive which comes after # will be better without this "sub-list" # reset current max to 0! current_max_sub_sum = 0 [1, -7, <2>, 15, -11, 2] most_seen = 2 current_max_sub_sum = 2 [1, -7, 2, <15>, -11, 2] most_seen = 17 current_max_sub_sum = 17 [1, -7, 2, 15, <-11>, 2] most_seen = 17 current_max_sub_sum = 6 # current sum went down, but not negative # no need to reset! [1, -7, 2, 15, -11, <2>] most_seen = 17 current_max_sub_sum = 8

Moving through the list **just once** is sufficient to find the maximum contiguous sub-sum.

### Instructions

**1.**

Write a function `max_sub_sum()`

which has a single parameter: `my_list`

.

`max_sub_sum()`

should return the maximum sum of contiguous values in `my_list`

.

**Your solution should run in O(N) time and O(1) space!**

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