The Maximum-Subarray Problem

Suppose that you have been offered the opportunity to invest in the Volatile Chemical Corporation. Like the chemicals the company produces, the stock price of the Volatile Chemical Corporation is rather volatile. You are allowed to buy one unit of stock only one time and then sell it at a later date, buying and selling after the close of trading for the day. To compensate for this restriction, you are allowed to learn what the price of the stock will be in the future. Your goal is to maximize your profit.

The problem 1: you are allowed to complete transaction only once, how do you calculate the max profit you may get?

def max_profit_1(prices):    max_profit = 0    n = len(prices)    if n > 1:        min_price = prices[0]        i = 1        while i < n:            if prices[i] > min_price:                profit = prices[i] - min_price                if profit > max_profit:                    max_profit = profit            else:                if min_price > prices[i]:                    min_price = prices[i]            i += 1    return max_profit

The problem 2: you are allowed to complete at most two transactions, how do you calculate the max profit you may get?

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

def max_profit_2(prices):    max_profit = 0    n = len(prices)    if n > 1:        left = [0 for i in range(n)]        right = [0 for i in range(n)]        min_price = prices[0]        i = 1        while i < n:            if prices[i] > min_price:                profit = prices[i] - min_price                if profit > max_profit:                    max_profit = profit            else:                min_price = prices[i]            left[i] = max_profit            i += 1        max_profit = 0        i = n - 2        max_price = prices[n - 1]        while i >= 0:            if prices[i] < max_price:                profit = max_price - prices[i]                if max_profit < profit:                    max_profit = profit            else:                max_price = prices[i]            right[i] = max_profit            i -= 1        i = 1        max_profit = left[n - 1]        while i < n - 2:            profit = left[i] + right[i + 1]            if profit > max_profit:                max_profit = profit            i += 1    return max_profit

The problem 3: you are allowed to complete transactions at most k (>2) times, how do you calculate the max profit you may get?

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

def max_profit_matrix(prices):    n = len(prices)    assert n > 1    ret = [[0 for i in range(n)] for j in range(n - 1)]    for i in range(n - 1):        min_price = prices[i]        max_profit = 0        j = i + 1        while j < n:            if prices[j] > min_price:                profit = prices[j] - min_price                if profit > max_profit:                    max_profit = profit            else:                min_price = prices[j]            ret[i][j] = max_profit            j += 1    return retdef max_profit_k(prices, k):    assert k > 2    max_profit = 0    n = len(prices)    if n > 1:        M = max_profit_matrix(prices)        cache = [[0 for i in range(n + 1)] for j in range(2)]        max_price = prices[n - 1]        max_profit = 0        i = n - 2        while i >= 0:            if prices[i] < max_price:                profit = max_price - prices[i]                if profit > max_profit:                    max_profit = profit            else:                max_price = prices[i]            cache[0][i] = max_profit            i -= 1        prev = 0        curr = 1        r = 2        while r <= k:            for i in range(n - 1):                j = i + 1                max_profit = cache[prev][i]                while j < n:                    profit = M[i][j] + cache[prev][j + 1]                    if profit > max_profit:                        max_profit = profit                    j += 1                cache[curr][i] = max_profit             prev = curr            curr = ((curr + 1) & 1)            r += 1        max_profit = cache[prev][0]    return max_profit

The test code:

if __name__ == "__main__":    prices = [100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97]    assert max_profit_1(prices) == 43    assert max_profit_2(prices) == 63    assert max_profit_k(prices, 3) == 81    assert max_profit_k(prices, 4) == 94