text: 
  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.
  
  
  
  
  
  
  
  
  
  
  // Dynamic Programming
  
  class Solution {
  public:
  
      int maxPUtil(vector<int> &prices, int i, int buyOrSell, int k, unordered_map<string, int> &dp){
          if(i>=prices.size() || k==0)
              return 0;
          string key=to_string(i)+to_string(buyOrSell)+to_string(k);
          if(dp.find(key)!=dp.end()) return dp[key];
          int res=0;
          if(buyOrSell==0){
              int buy=maxPUtil(prices, i+1, 1, k, dp)-prices[i];
              int noBuy=maxPUtil(prices, i+1, 0, k, dp);
              res+=max(buy, noBuy);
          } else {
              int sell=maxPUtil(prices, i+1, 0, k-1, dp)+prices[i];
              int noSell=maxPUtil(prices, i+1, 1, k, dp);
              res+=max(sell, noSell);
          }
          dp[key]=res;
          return res;
      }
  
      int maxProfit(vector<int>& prices) {
          unordered_map<string, int> dp;
          return maxPUtil(prices, 0, 0, 2, dp);  // <prices, i, buyOrSell, k, dp>     buy-->0   sell-->1
      }
  };
  
  
  
  
  
  
  
  
  
  
  
  
  
  // Recursive (TLE)
  
  class Solution {
  public:
  
      int maxPUtil(vector<int> &prices, int i, int buyOrSell, int k){
          if(i>=prices.size() || k==0)
              return 0;
          int res=0;
          if(buyOrSell==0){
              int buy=maxPUtil(prices, i+1, 1, k)-prices[i];
              int noBuy=maxPUtil(prices, i+1, 0, k);
              res+=max(buy, noBuy);
          } else {
              int sell=maxPUtil(prices, i+1, 0, k-1)+prices[i];
              int noSell=maxPUtil(prices, i+1, 1, k);
              res+=max(sell, noSell);
          }
          return res;
      }
  
      int maxProfit(vector<int>& prices) {
          return maxPUtil(prices, 0, 0, 2);  // <prices, i, buyOrSell, k>     buy-->0   sell-->1
      }
  };
  