Best Time to Buy and Sell Stock with Transaction Fee

Description

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

Example 1:

Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
Buying at prices[0] = 1
Selling at prices[3] = 8
Buying at prices[4] = 4
Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000.

0 < prices[i] < 50000.

0 <= fee < 50000.

Solution(javascript)

/*
 * @lc app=leetcode id=714 lang=javascript
 *
 * [714] Best Time to Buy and Sell Stock with Transaction Fee
 */

// @lc code=start
// /** 同时需要保证交易的次数尽可能的小
//  * Top-down: Maximum call stack size exceeded
//  * @param {number[]} prices
//  * @param {number} fee
//  * @return {number}
//  */
// const maxProfit = function (prices, fee) {
//   const BUY = 0
//   const SELL = 1
//   const memo = []
//   const aux = (index, prev) => {
//     memo[index] = memo[index] || []
//     if (memo[index][prev] !== undefined) {
//       return memo[index][prev]
//     }
//     if (index >= prices.length) {
//       return 0
//     }
//     if (prev === SELL) {
//       memo[index][prev] = Math.max(
//         -prices[index] + aux(index + 1, BUY),
//         aux(index + 1, SELL),
//       )
//     } else {
//       memo[index][prev] = Math.max(
//         prices[index] - fee + aux(index + 1, SELL),
//         aux(index + 1, BUY),
//       )
//     }
//     return memo[index][prev]
//   }
//   return aux(0, SELL)
// }
/**
 * Bottom-up
 * @param {number[]} prices
 * @param {number} fee
 * @return {number}
 */
const maxProfit = function (prices, fee) {
  const BUY = 0
  const SELL = 1
  let memo = [0, 0]
  for (let index = prices.length - 1; index >= 0; index--) {
    const current = []
    current[SELL] = Math.max(
      -prices[index] + memo[BUY],
      memo[SELL],
    )
    current[BUY] = Math.max(
      prices[index] - fee + memo[SELL],
      memo[BUY],
    )
    memo = current
  }
  return memo[SELL]
}
// @lc code=end