• 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 = [0] * (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]
``````