309. Best Time to Buy and Sell Stock with Cooldown

# Time:  O(n)
# Space: O(1)


class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        buy, sell, coolDown = [0] * 2, [0] * 2, [0] * 2
        buy[0] = -prices[0]
        for i in xrange(1, len(prices)):
            # Bought before or buy today.
            buy[i % 2] = max(buy[(i - 1) % 2],
                             coolDown[(i - 1) % 2] - prices[i])
            # Sell today.
            sell[i % 2] = buy[(i - 1) % 2] + prices[i]
            # Sold before yesterday or sold yesterday.
            coolDown[i % 2] = max(coolDown[(i - 1) % 2], sell[(i - 1) % 2])
        return max(coolDown[(len(prices) - 1) % 2],
                   sell[(len(prices) - 1) % 2])

Solution from kamyu104/LeetCode-Solutions · MIT