best-time-to-buy-and-sell-stock-ii.js

/**
 * Best Time to Buy and Sell Stock II
 *
 * Say you have an array for which the ith element is the price of a given stock on day i.
 *
 * Design an algorithm to find the maximum profit. You may complete as many transactions as you like
 * (i.e., buy one and sell one share of the stock multiple times).
 *
 * 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: [7,1,5,3,6,4]
 * Output: 7
 * Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
 *              Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
 *
 * Example 2:
 *
 * Input: [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: [7,6,4,3,1]
 * Output: 0
 * Explanation: In this case, no transaction is done, i.e. max profit = 0.
 */

/**
 * Solution I
 *
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit_I = prices => {
  if (!prices) {
    return 0;
  }

  let profit = 0;

  for (let i = 1; i < prices.length; i++) {
    profit += Math.max(prices[i] - prices[i - 1], 0);
  }

  return profit;
};

/**
 * Solution II
 *
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit_II = prices => {
  if (!prices) {
    return 0;
  }

  let cash = 0;
  let hold = -prices[0];

  for (let i = 1; i < prices.length; i++) {
    cash = Math.max(cash, hold + prices[i]);
    hold = Math.max(hold, cash - prices[i]);
  }

  return cash;
};

export { maxProfit_I, maxProfit_II };