122. Best Time to Buy and Sell Stock II

1. Description

You are given an integer array prices where prices[i] is the price of a given stock on the $i^{th}$ day.
On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can sell and buy the stock multiple times on the same day, ensuring you never hold more than one share of the stock.
Find and return the maximum profit you can achieve.

2. Example

Example 1

Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.

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.
Total profit is 4.

Example 3

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

3. Constraints

• 1 <= prices.length <= 3 * $10^4$
• 0 <= prices[i] <= $10^4$

4. Solutions

Dynamic Programming

n = prices.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int maxProfit(const vector<int> &prices) {
        int max_profit = 0;
        // a: max profit when holding a stock
        // b: max profit when not holding a stock
        int a1 = -prices[0], b1 = 0, a2 = 0, b2 = 0;
        for (int i = 1, n = prices.size(); i < n; ++i) {
            a2 = max(a1, b1 - prices[i]);
            b2 = max(b1, a1 + prices[i]);

            a1 = a2;
            b1 = b2;
            max_profit = max(b2, max_profit);
        }

        return max_profit;
    }
};

Greedy

n = prices.size()
Time complexity: O(n)
Space complexity: O(1)

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

        return max_profit;
    }
};