LeetCode-in-Ruby.github.io

121. Best Time to Buy and Sell Stock

Easy

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [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. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]

Output: 0

Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

Solution

# @param {Integer[]} prices
# @return {Integer}
def max_profit(prices)
  max_profit = 0
  min_price = prices[0]

  (1...prices.length).each do |i|
    if prices[i] > min_price
      max_profit = [max_profit, prices[i] - min_price].max
    else
      min_price = prices[i]
    end
  end

  max_profit
end

This site is open source. Improve this page.