> 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-iii.md).

# Best Time to Buy and Sell Stock III

## Problem Statement

<br>

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).

&#x20;

**Example 1:**

<pre><code><strong>Input: prices = [3,3,5,0,0,3,1,4]
</strong><strong>Output: 6
</strong><strong>Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
</strong>Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
</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>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.
</code></pre>

**Example 3:**

<pre><code><strong>Input: prices = [7,6,4,3,1]
</strong><strong>Output: 0
</strong><strong>Explanation: In this case, no transaction is done, i.e. max profit = 0.
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= prices.length <= 105`
* `0 <= prices[i] <= 105`\ <br>

## Intuition

```
Just introduce another state number of tries (i.e is 2)

Becomes 3D Dp and solve as same as Earlier question
```

### Links

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

### Video Links

<https://www.youtube.com/watch?v=-uQGzhYj8BQ&ab_channel=takeUforward>

### Approach 1:

```
Memoization
```

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

```cpp
class Solution {
public:
    int find_ans(vector<int>& prices, int index, bool buy, int tries, 
    vector<vector<vector<int>>> &dp){
        if(index == prices.size() or tries == 0)
            return 0;

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

        int profit;

        if(buy){
            profit = max( -prices[index] + find_ans(prices, index+1, false, tries, dp),
                            find_ans(prices, index+1, true, tries, dp) );
        }

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

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

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int tries = 2;
        vector<vector<vector<int>>> dp(n, vector<vector<int>> (2, vector<int>(3,-1)));

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

{% endcode %}

### Approach 2:

```
Tabulation
```

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

```cpp
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<vector<int>>> dp(n+1, vector<vector<int>> (2, vector<int>(3,0)));

        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                for(int tries=1; tries<=2; tries++){
                    // Starting tries from 1 not 0 because it is base case condition
                    //At tries = 0 we return 0 as cannot buy  
                    int profit;
                    if(buy){
                        profit = max( -prices[index] + dp[index+1][0][tries],
                                        dp[index+1][1][tries]);
                    }

                    else{
                        profit = max( +prices[index] + dp[index+1][1][tries-1],
                                        dp[index+1][0][tries]);
                    }

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

        return dp[0][1][2];
    }
};
```

{% endcode %}

### Approach 3:

```
Space Optimized
```

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

```cpp
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> cur(2, vector<int>(3,0));
        vector<vector<int>> prev(2, vector<int>(3,0));


        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                for(int tries=1; tries<=2; tries++){
                    // Starting tries from 1 not 0 because it is base case condition
                    //At tries = 0 we return 0 as cannot buy  
                    int profit;
                    if(buy){
                        profit = max( -prices[index] + prev[0][tries],
                                        prev[1][tries]);
                    }

                    else{
                        profit = max( +prices[index] + prev[1][tries-1],
                                       prev[0][tries]);
                    }

                    cur[buy][tries] = profit;
                }
            }
            prev = cur;
        }

        return prev[1][2];
    }
};
```

{% endcode %}

### Approach 4:

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

```cpp
Can maintain a N*4(Transaction number) State
Instead of N*2(Buy)*3(Transaction count)

In this 

0 1 2 3
B S B S

On even tries you buy and odd you sell

Base Cases will be 

At trasaction number == 4 we return 0 

```

{% endcode %}

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