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Best Time to Buy and Sell Stock III⚓︎

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Description⚓︎

You are given an array prices where prices[i] is the price of a given stock on the i-th 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:
1
2
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:
1
2
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 <= 10^5
  • 0 <= prices[i] <= 10^5

Solution⚓︎

Prefix-Suffix Decomposition⚓︎

See reference (Chinese).

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class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<int> f(n + 2);

        for (int i = 1, min_p = INT_MAX; i <= n; i++) {
            f[i] = max(f[i - 1], prices[i - 1] - min_p);
            min_p = min(min_p, prices[i - 1]);
        }

        int res = 0;
        for (int i = n, max_p = 0; i >= 1; i--) {
            res = max(res, max_p - prices[i - 1] + f[i - 1]);
            max_p = max(max_p, prices[i - 1]);
        }

        return res;
    }
};