content: package by.andd3dfx.dynamic;
  
  /**
   * <pre>
   * <a href="https://leetcode.com/explore/interview/card/top-interview-questions-easy/97/dynamic-programming/572/">Task description</a>
   *
   * 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.
   * </pre>
   *
   * @see <a href="https://youtu.be/XYPIKrEBbCc">Video solution</a>
   */
  public class BestTimeToBuyNSellStock {
  
      public int maxProfit_ON2(int[] prices) {
          var maxProfit = 0;
          for (int indexToBuy = 0; indexToBuy < prices.length - 1; indexToBuy++) {
              for (int indexToSell = indexToBuy + 1; indexToSell < prices.length; indexToSell++) {
                  var profit = prices[indexToSell] - prices[indexToBuy];
                  if (profit > maxProfit) {
                      maxProfit = profit;
                  }
              }
          }
          return maxProfit;
      }
  
      public int maxProfit_ON(int[] prices) {
          int minPrice = Integer.MAX_VALUE;
          int maxProfit = 0;
          for (int price : prices) {
              if (price < minPrice) {
                  minPrice = price;
              } else if (price - minPrice > maxProfit) {
                  maxProfit = price - minPrice;
              }
          }
          return maxProfit;
      }
  }
  