## # 123. Best Time to Buy and Sell Stock III

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 two transactions.

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

Example 1:

``````Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: 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: [1,2,3,4,5]
Output: 4
Explanation: 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: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
``````

### # Solution

Approach 1: DP. Make a transaction once by finding the best deal so far from the left and right, then find a split point where it maximizes `left[i] + right[i+1]`.

Approach 2: DP. Separate the first and second transactions.

### # Code (Python)

Approach 1:

``````    def maxProfit(self, prices: List[int]) -> int:
if len(prices) <= 1:
return 0
# make a transaction once, find the best deal so far from the left and right
# then find a split point where it maximizes left[i] + right[i+1]
left = 
lowest = prices
for i in range(1, len(prices)):
left.append(max(left[-1], prices[i] - lowest))
lowest = min(lowest, prices[i])
right =  * len(prices)
highest = prices[-1]
for i in range(len(prices) - 2, -1, -1):
right[i] = max(right[i+1], highest - prices[i])
highest = max(highest, prices[i])

max_profit = left[-1]
for i in range(len(prices) - 1):
max_profit = max(max_profit, left[i] + right[i+1])

return max_profit
``````

### # Code (C++)

Approach 1:

``````// Make a transaction once, find the best deal so far from the left and right
// then find a split point where it maximizes left[i] + right[i+1].
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if (n < 2)
return 0;
// From left to right.
int leftBuy = 0 - prices;
int leftSell[n];
leftSell = 0;
for (int i = 1; i < n; ++i)
{
leftSell[i] = std::max(leftSell[i-1], leftBuy + prices[i]);
}
// From right to left.
int rightBuy = 0 - prices[n-1];
int rightSell[n];
rightSell[n-1] = 0;
for (int i = n-2; i >= 0; --i)
{
rightSell[i] = std::min(rightSell[i+1], rightBuy + prices[i]);
}
// Get the max split point.
int maxVal = 0;
for (int i = 0; i < n; ++i)
{
maxVal = std::max(maxVal, leftSell[i] - rightSell[i]);
}
return maxVal;
}
};
``````

Approach 2:

``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if (n < 2)
return 0;
vector<int> sell1(n+1, 0);
vector<int> sell2(n+1, 0);
for (int i = 1; i <= n; ++i)
{
if (i > 1)
sell1[i] = std::max(sell1[i-1], buy1[i-1] + prices[i-1]);
if (i > 2)
if (i > 3)
sell2[i] = std::max(sell2[i-1], buy2[i-1] + prices[i-1]);
}
return std::max(sell1[n], sell2[n]);
}
};
``````
``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if (n < 2)
return 0;
int prevB1 = 0 - prices;
int prevS1 = 0;
int prevB2 = 0 - prices;
int prevS2 = 0;
for (int i = 1; i < n; ++i)
{
int currB1 = std::max(prevB1, 0 - prices[i]);
int currS1 = std::max(prevS1, prevB1 + prices[i]);
int currB2 = 0;
if (i > 1)
currB2 = std::max(prevB2, prevS1 - prices[i]);
int currS2 = 0;
if (i > 2)
currS2 = std::max(prevS2, prevB2 + prices[i]);
prevB1 = currB1;
prevS1 = currS1;
if (i > 1)
prevB2 = currB2;
if (i > 2)
prevS2 = currS2;
}
return std::max(prevS1, prevS2);
}
};
``````
``````class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if (n < 2)
return 0;
int res = 0;