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

/*
 * @lc app=leetcode id=121 lang=javascript
 *
 * [121] Best Time to Buy and Sell Stock
 *
 * https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
 *
 * algorithms
 * Easy (50.41%)
 * Likes:    5552
 * Dislikes: 242
 * Total Accepted:    920.6K
 * Total Submissions: 1.8M
 * Testcase Example:  '[7,1,5,3,6,4]'
 *
 * Say you have an array for which the i^th element is the price of a given
 * stock on day i.
 *
 * If you were only permitted to complete at most one transaction (i.e., buy
 * one and sell one share of the stock), design an algorithm to find the
 * maximum profit.
 *
 * Note that you cannot sell a stock before you buy one.
 *
 * Example 1:
 *
 *
 * Input: [7,1,5,3,6,4]
 * Output: 5
 * Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit
 * = 6-1 = 5.
 * Not 7-1 = 6, as selling price needs to be larger than buying price.
 *
 *
 * Example 2:
 *
 *
 * Input: [7,6,4,3,1]
 * Output: 0
 * Explanation: In this case, no transaction is done, i.e. max profit = 0.
 *
 *
 */

// @lc code=start
// /**
//  * @param {number[]} prices
//  * @return {number}
//  */
// var maxProfit = function (prices) {
//     let max = 0;
//     let left = prices.slice();
//     for (let i in prices) {
//         left.shift();
//         let currentMax = Math.max(...left) - prices[i];
//         max = Math.max(max, currentMax);
//     }
//     return max;
// };
/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
    let minPrice = Math.max(...prices);
    let maxMoney = 0;
    for (let price of prices) {
        if (price < minPrice) minPrice = price;
        else if (price - minPrice > maxMoney)
            maxMoney = price - minPrice;
    }
    return maxMoney;
};
// @lc code=end