123. Best Time to Buy and Sell Stock III

Hard (Khó) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)
⏱️ Thời gian O(n)
💾 Không gian O(n)

💻 Lời Giải

C++ 0123-best-time-to-buy-and-sell-stock-iii.cpp 
const int MAXN = 1e5;
int memo[MAXN][2][3];

class Solution {
private:
    vector<int> prices;
    int n;
        
public:
    int dp(int i, int state, int cnt) {
        if (i == n) {
            return 0;
        }
        if (memo[i][state][cnt] != -1) {
            return memo[i][state][cnt];
        }
        if (cnt == 2) {
            return memo[i][state][cnt] = dp(i + 1, state, cnt);
        }
        if (state == 0) {
            return memo[i][state][cnt] = max(
                -prices[i] + dp(i + 1, !state, cnt), 
                dp(i + 1, state, cnt)
            );
        }
        else {
            return memo[i][state][cnt] = max(
                prices[i] + dp(i + 1, !state, cnt + 1), 
                dp(i + 1, state, cnt)
            );
        }
    }
    
    int maxProfit(vector<int>& prices) {
        memset(memo, -1, sizeof(memo));
        this->prices = prices;
        this-> n = prices.size();
        return dp(0, 0, 0);
    }
};