188. Best Time to Buy and Sell Stock IV

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).

 

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

 

Solutions

Solution: Dynamic Programming

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

 

JavaScript

/**
 * @param {number} k
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit = function (k, prices) {
  const n = prices.length;

  if (k >= n / 2) {
    let result = 0;

    for (let index = 0; index < n - 1; index++) {
      const current = prices[index];
      const next = prices[index + 1];

      if (current >= next) continue;
      result += next - current;
    }
    return result;
  }
  const hold = new Array(k + 1).fill(Number.MIN_SAFE_INTEGER);
  const sell = new Array(k + 1).fill(0);

  for (const price of prices) {
    for (let times = 1; times <= k; times++) {
      sell[times] = Math.max(sell[times], hold[times] + price);
      hold[times] = Math.max(hold[times], sell[times - 1] - price);
    }
  }
  return sell[k];
};