Best Time to Buy and Sell Stock II

Problem Statement

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

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.

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.
Total profit is 4.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints:

1 <= prices.length <= 3 * 104
0 <= prices[i] <= 104

Intuition

Traverse and if i - i-1 th index gives profit, then take it

Links

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

Video Links

https://www.youtube.com/watch?v=nGJmxkUJQGs&ab_channel=takeUforward

Approach 1:

Iterate over array

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int profit = 0;

        for(int i=1; i<prices.size(); i++){
            if( prices[i]-prices[i-1] >0 )
                profit += prices[i]-prices[i-1];
        }

        return profit;
    }
};

Approach 2:

Memoization

Maintained two states- index, buy/sell 

    If buy, 
        two cases, buy/dont buy change flag accordingly
        
     else if sell
        two cases, Sell/Not sell

C++

class Solution {
public:
    int find_ans(vector<int>& prices, int index, bool buy, vector<vector<int>> &dp){

        if(index == prices.size())
            return 0;

        if(dp[index][buy] != -1)
            return dp[index][buy];

        int profit;

        if(buy){
            profit = max ( -prices[index] + find_ans(prices, index+1, false, dp), 
                                find_ans(prices, index+1, true, dp) );
            /*
                If buy, 
                    two cases, buy/dont buy change flag accordingly
                    
                 else if sell
                    two cases, Sell/Not sell 
            */
        }

        else{
            profit = max( prices[index] + find_ans(prices, index+1, true, dp) ,
                                    find_ans(prices, index+1, false, dp) );
        }

        return dp[index][buy] = profit;
    }

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp (n, vector<int>(2,-1));

        return find_ans(prices, 0, true, dp);
    }
};

Approach 3:

Tabulation

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp (n+1, vector<int>(2,0));

        //Base Case, No need here just added
        dp[n][0] = dp[n][1] = 0;

        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                int profit;

                if(buy)
                    profit = max ( -prices[index] + dp[index+1][0], dp[index+1][1] );
                
                else
                    profit = max( prices[index] + dp[index+1][1] ,dp[index+1][0] );

                dp[index][buy] = profit;
            }
        }

        return dp[0][1];
    }
};

Approach 4:

Space Optimization

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<int> prev(2,0), cur(2,0);

        //Base Case, No need here just added
        cur[0] = cur[1] = 0;

        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                int profit;

                if(buy)
                    profit = max ( -prices[index] + cur[0], cur[1] );
                
                else
                    profit = max( prices[index] + cur[1] ,cur[0] );

                prev[buy] = profit;
            }

            cur = prev;
        }

        return prev[1];
    }
};