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
Best Time to Buy and Sell Stock IV Solutions
✅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]
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