121. Best Time to Buy and Sell Stock.md

Difficulty	Related Topics	Similar Questions
Easy	Array Dynamic Programming https://leetcode.com/tag/array https://leetcode.com/tag/dynamic-programming	Maximum Subarray Best Time to Buy and Sell Stock II Best Time to Buy and Sell Stock III Best Time to Buy and Sell Stock IV Best Time to Buy and Sell Stock with Cooldown https://leetcode.com/problems/maximum-subarray https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown

121. Best Time to Buy and Sell Stock

Problem:

Say you have an array for which the ith 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.

Solution:

Only care about positive profits. Take the frist item as base and scan to the right. If we encounter an item j whose price price[j] is lower than the base (which means if we sell now the profit would be negative), we sell j-1 instead and make j the new base.

Because price[j] is lower that the base, using j as new base is guaranteed to gain more profit comparing to the old one.

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
  let max = 0
  let base = prices[0]
  for (let i = 1; i < prices.length; i++) {
    const profit = prices[i] - base
    if (profit > max) {
      max = profit
    } else if (profit < 0) {
      base = prices[i]
    }
  }
  return max
};

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