# Best Time to Buy and Sell Stock IV LeetCode Solution Share:

Best Time to Buy and Sell Stock IV 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.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

```Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
```

Example 2:

```Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
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.
```

Constraints:

• `0 <= k <= 100`
• `0 <= prices.length <= 1000`
• `0 <= prices[i] <= 1000`

Time: O(nk)
Space: O(k)

### C++

``````class Solution {
public:
int maxProfit(int k, vector<int>& prices) {
if (k >= prices.size() / 2) {
int sell = 0;
int hold = INT_MIN;

for (const int price : prices) {
sell = max(sell, hold + price);
hold = max(hold, sell - price);
}

return sell;
}

vector<int> sell(k + 1);
vector<int> hold(k + 1, INT_MIN);

for (const int price : prices)
for (int i = k; i > 0; --i) {
sell[i] = max(sell[i], hold[i] + price);
hold[i] = max(hold[i], sell[i - 1] - price);
}

return sell[k];
}
};
``````

### Java

`````` class Solution {
public int maxProfit(int k, int[] prices) {
if (k >= prices.length / 2) {
int sell = 0;
int hold = Integer.MIN_VALUE;

for (final int price : prices) {
sell = Math.max(sell, hold + price);
hold = Math.max(hold, sell - price);
}

return sell;
}

int[] sell = new int[k + 1];
int[] hold = new int[k + 1];
Arrays.fill(hold, Integer.MIN_VALUE);

for (final int price : prices)
for (int i = k; i > 0; --i) {
sell[i] = Math.max(sell[i], hold[i] + price);
hold[i] = Math.max(hold[i], sell[i - 1] - price);
}

return sell[k];
}
}

``````

### Python

``````
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
if k >= len(prices) // 2:
sell = 0
hold = -math.inf

for price in prices:
sell = max(sell, hold + price)
hold = max(hold, sell - price)

return sell

sell =  * (k + 1)
hold = [-math.inf] * (k + 1)

for price in prices:
for i in range(k, 0, -1):
sell[i] = max(sell[i], hold[i] + price)
hold[i] = max(hold[i], sell[i - 1] - price)

return sell[k]
``````

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