**Best Time to Buy and Sell Stock IV** **LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++**

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** Link for the Problem** – Best Time to Buy and Sell Stock IV– LeetCode Problem

Best Time to Buy and Sell Stock IV– LeetCode Problem

**Problem:**

You are given an integer array `prices`

where `prices[i]`

is the price of a given stock on the `i`

day, and an integer ^{th}`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:2Explanation: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:7Explanation: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– LeetCode Solutions

Best Time to Buy and Sell Stock IVSolution in 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]; } };

Best Time to Buy and Sell Stock IVSolution in 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]; } }

Best Time to Buy and Sell Stock IVSolution in 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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