Best Time to Buy and Sell Stock II

Say you have an array prices for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [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.

Example 2:

Input: [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: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:

• 1 <= prices.length <= 3 * 10 ^ 4

• 0 <= prices[i] <= 10 ^ 4

class Solution {
    public int maxProfit(int[] prices) {
        int T_ik0 = 0, T_ik1 = Integer.MIN_VALUE;

        for (int price : prices) {
            int T_ik0_old = T_ik0;
            // If we want 0 stocks in our hand at end of ith day
            // Either take ans from 0 stock i-1th day
            // Or take ans from 1 stock i-1th day and sell it
            T_ik0 = Math.max(T_ik0, T_ik1 + price);
            // If we want 1 stock in our hand at end of ith day
            // Either take ans from 1 stock i-1th day
            // Or take ans from 0 stock i-1th day and but 1 stock at ith day
            T_ik1 = Math.max(T_ik1, T_ik0_old - price);
        }

        return T_ik0;
    }
}