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LC-0123 Hard LeetCode

123. Best Time to Buy and Sell Stock III

Read the full problem statement on LeetCode.
Difficulty: hard Acceptance: 51% Topics: Array, Dynamic Programming
View full problem on LeetCode
Reference solution (spoiler · python)
# Time:  O(n)
# Space: O(1)

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        hold1, hold2 = float("-inf"), float("-inf")
        release1, release2 = 0, 0
        for i in prices:
            hold1 = max(hold1, -i)
            release1 = max(release1, hold1 + i)
            hold2 = max(hold2, release1 - i)
            release2 = max(release2, hold2 + i)
        return release2


# Time:  O(k * n)
# Space: O(k)
class Solution2(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        def maxAtMostKPairsProfit(prices, k):
            max_buy = [float("-inf") for _ in xrange(k + 1)]
            max_sell = [0 for _ in xrange(k + 1)]
            for i in xrange(len(prices)):
                for j in xrange(1, k + 1):
                    max_buy[j] = max(max_buy[j], max_sell[j-1] - prices[i])
                    max_sell[j] = max(max_sell[j], max_buy[j] + prices[i])
            return max_sell[k]

        return maxAtMostKPairsProfit(prices, 2)


# Time:  O(n)
# Space: O(n)
class Solution3(object):
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        min_price, max_profit_from_left, max_profits_from_left = \
            float("inf"), 0, []
        for price in prices:
            min_price = min(min_price, price)
            max_profit_from_left = max(max_profit_from_left, price - min_price)
            max_profits_from_left.append(max_profit_from_left)

        max_price, max_profit_from_right, max_profits_from_right = 0, 0, []
        for i in reversed(range(len(prices))):
            max_price = max(max_price, prices[i])
            max_profit_from_right = max(max_profit_from_right,
                                        max_price - prices[i])
            max_profits_from_right.insert(0, max_profit_from_right)

        max_profit = 0
        for i in range(len(prices)):
            max_profit = max(max_profit,
                             max_profits_from_left[i] +
                             max_profits_from_right[i])

        return max_profit

Solution from kamyu104/LeetCode-Solutions · MIT