Problem Description
Given an array of stock prices where prices[i] is the price on the i-th day, determine the maximum profit that can be achieved. You can buy and sell multiple times; however, you can hold at most one share at any time. Transactions on the same day (buy and sell on the same day) are allowed.
Key Insights
- You can make as many transactions as needed as long as you are not holding more than one share.
- The best strategy is to capture every upward movement in price.
- Instead of finding the best pair of days for each transaction, sum the differences between consecutive days that indicate an increase.
- This greedy approach avoids complex dynamic programming structures while still ensuring maximum profit.
Space and Time Complexity
Time Complexity: O(n), where n is the length of the prices array, since we traverse the list once. Space Complexity: O(1), as we only need a few integer variables for tracking profit.
Solution
The solution uses a greedy approach by iterating over the prices list and adding up profits each time there is a rise from one day to the next. If prices[i] is lower than prices[i+1], buying on day i and selling on day i+1 will add to the overall profit. The idea is that accumulating all these small profits yields the maximum profit possible by making as many transactions as necessary. The approach avoids explicit tracking of all transactions, requiring only constant extra space.