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Best Time to Buy and Sell Stock IV

Problem Description​

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).

Examples​

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​

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

Approach​

Approach to Find Maximum Profit with at Most k Transactions​

  1. Edge Case Handling:

    • If k is greater than or equal to half the length of prices, it means we can perform an unlimited number of transactions.
    • In this case, initialize sell to 0 (maximum profit with no stock) and hold to negative infinity (maximum profit with stock).
    • Iterate through prices to update sell and hold based on the maximum possible profit at each price.
    • Return the final sell value as the result.

  2. General Case:

    • If k is less than half the length of prices, initialize two lists: sell to store maximum profit after selling a stock, and hold to store maximum profit after buying a stock.
    • Both lists have a length of k + 1, with sell initialized to 0 and hold initialized to negative infinity.

  3. Iterate Through Prices:

    • For each price in prices, iterate from k down to 1 to update sell and hold.
    • Update sell[i] to be the maximum of itself or the profit from selling the stock held at the current price.
    • Update hold[i] to be the maximum of itself or the profit from buying a stock at the current price.

  4. Return Results: Return sell[k], which represents the maximum profit with at most k transactions.

Solution​

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];
  }
};

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]

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];
  }
}

Complexity analysis​

  • Time Complexity: 𝑂(nk)𝑂(nk)
  • Space Complexity: O(k)O(k)