Best Time to Buy and Sell Stock III.cpp

/*
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: prices = [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: prices = [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: prices = [7,6,4,3,1]
Output: 0
Explanation: 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
*/

class Solution
{
public:
  int maxProfit(vector<int> &nums)
  {
    int n = nums.size();
    if (n <= 1)
      return 0;
    vector<int> profit(n, 0);

    int max_so_far = nums[n - 1];
    for (int i = n - 2; i >= 0; --i)
    {
      max_so_far = max(max_so_far, nums[i]);
      profit[i] = max(profit[i + 1], max_so_far - nums[i]);
    }

    int min_so_far = nums[0];
    for (int i = 1; i < n; ++i)
    {
      min_so_far = min(min_so_far, nums[i]);
      profit[i] = max(profit[i - 1], profit[i] + nums[i] - min_so_far);
    }

    return profit[n - 1];
  }
};