max_profit.py

"""
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).
Elements of input arrays can be modified.
"""

import sys


def solution(A):
    min_price = sys.maxsize
    max_profit = 0
    for a in A:
        min_price = min([min_price, a])
        max_profit = max([max_profit, a - min_price])
    return max_profit