動機

用算k個的dp就可以解這題，但是想介紹當初沒有用dp的解法

Problem

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

 

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]Output: 6Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]Output: 4Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]Output: 0Explanation: In this case, no transaction is done, i.e. max profit = 0.

Example 4:

Input: prices = [1]Output: 0

 

Constraints:

  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 105

Sol

最多只做兩次，所以可以算只做一次的最大與兩次的最大

兩次的最大透過

  1. 從左看過去到第i天最大的交易
  2. 從右看過去到第i天最大的交易

相加

# @param {Integer[]} prices
# @return {Integer}
# k = 1
def f(prices)
    lowest_price = prices[0]
    ans = 0
    i = 1
    while i < prices.size
        lowest_price = [lowest_price,prices[i]].min
        ans = [ans, prices[i]-lowest_price].max
        i+=1
    end
    return ans
end
def max_profit(ps)
    fore = ps.clone
    back = ps.clone
    back[-1] = back[0] = 0
    fore[-1] = fore[0] = 0
    small = ps[0]
    (1...ps.size).each do |i|
        fore[i] = ps[i]-small > fore[i-1]? ps[i]-small : fore[i-1]
        small = [small, ps[i]].min
        
    end
    big = ps[-1]
    (ps.size-2).downto(1).each do |i|
        back[i] = big-ps[i]>back[i+1]? big-ps[i] : back[i+1]
        big = [big, ps[i]].max
        
    end
    
    p fore
    p back
    back.shift
    back.push 0
    ans = 0
    (0...ps.size).each do |i|
        ans = [ans, back[i]+fore[i]].max
    end
    return [ans, f(ps)].max
end