Best Time to Buy and Sell Stock IV

Question

Say you have an array 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 k transactions.

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

Tags

• Array
• Dynamic Programming

Thought

As we have discussed the universal solution in Best Time to Buy and Sell Stock III, we will use that code directly. The only problem is that the value of k is very large in some test case, which might harm the memory efficiency. Thus, we add a check for the value of k in the code to avoid the waste of memory.

Code

class Solution(object):
    def maxProfit(self, k, prices):
        """

        global[i][j]=max(local[i][j],global[i-1][j])，

        local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff)

        :type k: int
        :type prices: List[int]
        :rtype: int
        """
        length = len(prices)
        if length == 0:
            return 0
        # k is big enougth to cover all ramps.
        if k >= length / 2:
            return sum(i - j
                       for i, j in zip(prices[1:], prices[:-1]) if i - j > 0)
        local_profit = [0] * (k+1)
        global_profit = [0] * (k+1)
        for i in xrange(length - 1):
            diff = prices[i + 1] - prices[i]
            for j in xrange(k, 0, -1):
                local_profit[j] = max(global_profit[j - 1] + max(diff, 0), local_profit[j] + diff)
                global_profit[j] = max(local_profit[j], global_profit[j])
        return global_profit[k]