Best Time to Buy and Sell Stock II | Leetcode 122 Solution

 122 . Best Time to Buy and Sell Stock II

   Memoization approach : 

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: 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.

 

Constraints:

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

solution :

  Explanation: x = 0 means we are going to buy stock. since we have two choices we can buy a particular stock or buy stock from the remaining array.

so, I wrote the code :

if(x == 0)
        {
            return dp[i][x] = max(-prices[i] + solve(prices, i+1, size, 1, dp) , solve(prices , i+1 , size , 0 , dp));
        }

it means if we select that index and then change x = 1, that is next time we are going to sell it at a different price, and the second choice is we will buy from the remaining array. 

another return statement is for sell of the stock. this will also work in a similar way.

class Solution { public:     int solve(vector<int>& prices , int i , int size , int x , vector<vector<int>>&dp)     {         if(i == size )         {             return 0;         }         if(dp[i][x] != -1)         {             return dp[i][x];         }         if(x == 0)         {             return dp[i][x] = max(-prices[i] + solve(prices, i+1, size, 1, dp) , solve(prices , i+1 , size , 0 , dp));         }                   return dp[i][x] = max(prices[i] + solve(prices , i+1 , size , 0 , dp), solve(prices , i+1, size , 1 , dp));     }     int maxProfit(vector<int>& prices) {         int size = prices.size();         int i =0;         int x = 0;         vector<vector<int>>dp(size+1 , vector<int>(2 , -1));         return solve(prices , i, size , x, dp);           } };