(Lesson 9, Maximum slice problem) MaxProfit

Problem

An 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 an 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.

Write an efficient algorithm for the following assumptions:

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

My solution

def solution(A): # write your code in Python 3.6 if len(A) == 0: return 0 start = 0 end = len(A)-1 min_val = A[start] max_val = A[end] while start <= end: min_val = min(min_val, A[start]) max_val = max(max_val, A[end]) gap_l = min_val - A[start+1] gap_r = A[end-1] - max_val if gap_l > gap_r: start += 1 else: end -= 1 return max_val-min_val