188. Best Time to Buy and Sell Stock IV

Problem

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

Solution

/**
 * @param {number} k
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(k, prices) {
  if (!prices.length) return 0
  let len = prices.length,
    res = 0
  if (k >= ~~(len / 2)) {
    for (let i = 1; i < len; i++) {
      res += Math.max(prices[i] - prices[i - 1], 0)
    }
    return res
  }
  const buy = new Array(k + 1).fill(Number.MIN_SAFE_INTEGER)
  const sell = new Array(k + 1).fill(0)

  for (let p of prices) {
    for (let i = 1; i <= k; i++) {
      buy[i] = Math.max(sell[i - 1] - p, buy[i])
      sell[i] = Math.max(buy[i] + p, sell[i])
    }
  }
  return sell[k]
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).