# 188. Best Time to Buy and Sell Stock IV

## Problem Statement

\


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

Find the maximum profit you can achieve. You may complete at most `k` transactions: i.e. you may buy at most `k` times and sell at most `k` times.

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

**Example 2:**

<pre><code><strong>Input: k = 2, prices = [3,2,6,5,0,3]
</strong><strong>Output: 7
</strong><strong>Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
</strong></code></pre>

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**Constraints:**

* `1 <= k <= 100`
* `1 <= prices.length <= 1000`
* `0 <= prices[i] <= 1000`\
  \


## Intuition

```
Exactly same as III Problem just tries=2 is changed to k
```

### Links

[https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)

### Video Links

[https://www.youtube.com/watch?v=IV1dHbk5CDc\&ab\_channel=takeUforward](https://www.youtube.com/watch?v=IV1dHbk5CDc\&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(int k, vector<int>& prices) {
        int n = prices.size();
        int tries = k;
        vector<vector<vector<int>>> dp(n, vector<vector<int>> (2, vector<int>(k+1,-1)));

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

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}

### Approach 4:

{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}

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###

