Say you have an array, A, 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 at most 2 transactions.Return the maximum possible profit.Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).Problem Constraints1 <= length(A) <= 7e51 <= A[i] <= 1e7Sample input57 2 4 8 7Sample output6

The problem can be solved using dynamic programming. Here is a step-by-step guide to the algorithm:

1. Initialize two arrays, profit1 and profit2, of the same length as the input array A. These arrays will store the maximum profit that can be made with at most one transaction and at most two transactions, respectively.

2. Set profit1[0] to 0, because no profit can be made on the first day. For each day i from 1 to the end of A, calculate profit1[i] as the maximum of profit1[i-1] (the maximum profit made up to the previous day) and A[i] - min_price, where min_price is the minimum price seen so far. This represents the maximum profit that can be made with a single transaction up to day i.

3. Set profit2[0] and profit2[1] to 0, because no profit can be made on the first two days with two transactions. For each day i from 2 to the end of A, calculate profit2[i] as the maximum of profit2[i-1] (the maximum profit made up to the previous day with two transactions) and A[i] - min_price_with_profit, where min_price_with_profit is the minimum price seen so far minus the profit that can be made with one transaction up to that day. This represents the maximum profit that can be made with two transactions up to day i.

4. The maximum profit that can be made with at most two transactions is profit2[-1], the last element of profit2.

Here is a Python implementation of the algorithm:

def maxProfit(A):
    if not A:
        return 0

    profit1 = [0] * len(A)
    profit2 = [0] * len(A)

    min_price = A[0]
    for i in range(1, len(A)):
        min_price = min(min_price, A[i])
        profit1[i] = max(profit1[i-1], A[i] - min_price)

    min_price_with_profit = A[0] - profit1[0]
    for i in range(2, len(A)):
        min_price_with_profit = min(min_price_with_profit, A[i] - profit1[i-1])
        profit2[i] = max(profit2[i-1], A[i] - min_price_with_profit)

    return profit2[-1]

For the sample input A = [5, 7, 2, 4, 8, 7], the function maxProfit(A) returns 6, which is the maximum profit that can be made with at most two transactions.