> For the complete documentation index, see [llms.txt](https://coding-9.gitbook.io/untitled/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://coding-9.gitbook.io/untitled/dynamic-programming/dp-on-stocks/best-time-to-buy-and-sell-stock-ii.md).

# Best Time to Buy and Sell Stock II

## Problem Statement

<br>

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*.

&#x20;

**Example 1:**

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

**Example 2:**

<pre><code><strong>Input: prices = [1,2,3,4,5]
</strong><strong>Output: 4
</strong><strong>Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
</strong>Total profit is 4.
</code></pre>

**Example 3:**

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

&#x20;

**Constraints:**

* `1 <= prices.length <= 3 * 104`
* `0 <= prices[i] <= 104`\ <br>

## 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
```

{% code title="C++" lineNumbers="true" %}

```cpp
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;
    }
};
```

{% endcode %}

### 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 
```

{% code title="C++" lineNumbers="true" %}

```cpp
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);
    }
};
```

{% endcode %}

### Approach 3:

```
Tabulation    
```

{% code title="C++" lineNumbers="true" %}

```cpp
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];
    }
};
```

{% endcode %}

### Approach 4:

```
Space Optimization
```

{% code title="C++" lineNumbers="true" fullWidth="false" %}

```cpp
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];
    }
};
```

{% endcode %}

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