Sharing an answer code of mine about MaxProfit problem of Codility lesson 9.

# Lesson 9: MaxProfit

A zero-indexed array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].

For example, consider the following array A consisting of six elements such that:

\[A[0] = 23171\\ A[1] = 21011\\ A[2] = 21123\\ A[3] = 21366\\ A[4] = 21013\\ A[5] = 21367\]If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.

Write a function,

```
def solution(A)
```

that, given a zero-indexed array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible to gain any profit.

For example, given array A consisting of six elements such that:

\[A[0] = 23171\\ A[1] = 21011\\ A[2] = 21123\\ A[3] = 21366\\ A[4] = 21013\\ A[5] = 21367\]the function should return 356, as explained above.

Assume that:

- N is an integer within the range [0..400,000];
- each element of array A is an integer within the range [0..200,000].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

# Answer Code in Python 3

- Time complexity: \(O(N)\)
- Space complexity: \(O(N)\)

```
# Time Complexity: O(N)
# Space Complexity: O(N)
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
# write your code in Python 3.6
std, max_pf = 0, 0
for i in range(len(A)):
if A[std] >= A[i]: std = i
else: max_pf = max(max_pf, A[i]-A[std])
return max_pf
```