Best Time to Buy and Sell Stock III

Description

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

 

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Example 4:

Input: prices = [1]
Output: 0

 

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 105

Solution(javascript)

/*
 * @lc app=leetcode id=123 lang=javascript
 *
 * [123] Best Time to Buy and Sell Stock III
 */

// @lc code=start
/**
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit = function (prices) {
  const memo = []
  const aux = (index, count) => {
    memo[index] = memo[index] || []
    if (memo[index][count] !== undefined) {
      return memo[index][count]
    }
    if (count > 4 || index > prices.length - 1) {
      return 0
    }
    if (count % 2 === 1) {
      memo[index][count] = Math.max(
        -prices[index] + aux(index + 1, count + 1),
        aux(index + 1, count),
      )
    } else {
      memo[index][count] = Math.max(
        prices[index] + aux(index + 1, count + 1),
        aux(index + 1, count),
      )
    }
    return memo[index][count]
  }
  return aux(0, 1)
}
// @lc code=end