Best Time to Buy and Sell Stock III 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
Best Time to Buy and Sell Stock III Solutions
✅Time: O(n)
✅Space: O(1)
C++
class Solution {
public:
int maxProfit(vector<int>& prices) {
int sellTwo = 0;
int holdTwo = INT_MIN;
int sellOne = 0;
int holdOne = INT_MIN;
for (const int price : prices) {
sellTwo = max(sellTwo, holdTwo + price);
holdTwo = max(holdTwo, sellOne - price);
sellOne = max(sellOne, holdOne + price);
holdOne = max(holdOne, -price);
}
return sellTwo;
}
};
Java
class Solution {
public int maxProfit(int[] prices) {
int sellTwo = 0;
int holdTwo = Integer.MIN_VALUE;
int sellOne = 0;
int holdOne = Integer.MIN_VALUE;
for (final int price : prices) {
sellTwo = Math.max(sellTwo, holdTwo + price);
holdTwo = Math.max(holdTwo, sellOne - price);
sellOne = Math.max(sellOne, holdOne + price);
holdOne = Math.max(holdOne, -price);
}
return sellTwo;
}
}
Python
class Solution:
def maxProfit(self, prices: List[int]) -> int:
sellTwo = 0
holdTwo = -math.inf
sellOne = 0
holdOne = -math.inf
for price in prices:
sellTwo = max(sellTwo, holdTwo + price)
holdTwo = max(holdTwo, sellOne - price)
sellOne = max(sellOne, holdOne + price)
holdOne = max(holdOne, -price)
return sellTwo
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