You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times). Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
1 <= prices.length <= 3 * 10^4
0 <= prices[i] <= 10^4
For the DP approach, we should track the local profit which is the difference between each two consecutive prices and this value can be added to the total profit if it is positive.. This solution would have a time complexity of O(N) and space complexity of O(1) where N is the size of prices array.
def maxProfit(prices):
"""
:type prices: List[int]
:rtype: int
:Time Complexity: O(N)
:Space Complexity: O(1)
"""
total_profit = 0
for i in range(1, len(prices)):
local_profit = prices[i] - prices[i - 1]
total_profit = max(total_profit, local_profit + total_profit)
return total_profit