Best Time to Buy and Sell Stock III
Problem Description
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 at most two transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
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.
Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Constraints:
- 1 <= prices.length <= 105
- 0 <= prices[i] <= 105
Solve it on LeetCode
Approaches
1. Brute Force
The brute force approach would involve checking every possible combination of two transactions (buy-sell pairs) to find the maximum profit. However, this approach is highly inefficient since it would require iterating over all possible pairs of transactions.
Complexity Analysis:
- Time Complexity: O(N^4), where N is the number of days. This is because the approach involves checking each pair of days for both transactions, resulting in a nested iteration.
- Space Complexity: O(1), no extra space is used aside from a few variables.
2. Dynamic Programming with State Machine
Intuition:
- We need an optimal way to calculate profits with at most two transactions.
- We define states based on whether we have completed transactions or holding stocks.
State Definitions:
0: before any transaction,1: holding stock after the first buy,2: completed the first sell,3: holding stock after the second buy,4: completed the second sell.
Each day, we determine the best action based on states defined above.
Code:
- Time Complexity: O(N), iterating over prices array once.
- Space Complexity: O(N), space used by dp array to store intermediate results.
3. Optimized Dynamic Programming
Intuition:
- We can reduce the space complexity by using only two variables for each state since each state depends only on the previous day.
Code:
- Time Complexity: O(N), single iteration through the prices array.
- Space Complexity: O(1), space used is constant as we only store state for the previous day.