123. Best Time to Buy and Sell Stock III.cpp

// 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



class Solution {
public:
    int maxProfit(vector<int>& prices) {
		vector<int> L, R; L = R = vector<int>(prices.size(), 0);
		int min_a = prices[0];
		for (int i = 1; i < prices.size(); i++) {
			L[i] = max(L[i - 1], prices[i] - min_a);
			min_a = min(min_a, prices[i]);
		}

		int max_a = prices.back();
		for (int i = prices.size() - 2; i >= 0; i--) {
			R[i] = max(R[i + 1], max_a - prices[i]);
			max_a = max(max_a, prices[i]);
		}

		int ans = R[0];
		for (int i = 1; i + 1 < prices.size(); i++)
			ans = max(ans, L[i] + R[i + 1]);
		return ans;	
    }
};