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:
If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A − A = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A − A = 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,
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:
the function should return 356, as explained above.
- N is an integer within the range [0..400,000];
- each element of array A is an integer within the range [0..200,000].
- 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:
- Space complexity:
# 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