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

Design an algorithm to find the maximum profit. You may complete at most two transactions.

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: [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: [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.

Solution:

func maxProfit(prices []int) int {
    k := 2
    lenN := len(prices)
    dp := make([][][]int, 2)// X day, Y times, Z hold stock or not 
    for i:= 0; i <= 1; i++ {
        dp[i] = make([][]int, k+1)
        for j:=0; j <= k; j++ {
            dp[i][j] = make([]int, 2)
        }
    }
    
    MIN_INFIN := math.MinInt32 //math.MinInt32
    for i:= 0; i <= k; i++ {
        dp[0][i][0] = 0                 
        dp[0][i][1] = MIN_INFIN
    }

    for i:= 1; i <= lenN; i++ {
        for j := 1;  j <= k; j++ {
            dp[i%2][j][0] = max(dp[(i-1)%2][j][0], dp[(i-1)%2][j][1] + prices[i-1]) // no stock today
            dp[i%2][j][1] = max(dp[(i-1)%2][j-1][0] - prices[i-1], dp[(i-1)%2][j][1]) // have stock today
        }
    }
    maxResult := 0;
    for i:= 0; i <= k; i++ {
        maxResult = max(maxResult, dp[(lenN)%2][i][0])
    }
    return maxResult
}

func max(a, b int) int {
    if a >= b {
        return a
    }
    return b
}