Leetcode Solution - Best Time to Buy and Sell Stock

July 06, 2021

Problem Statement
Brute Force - O(n^2)
Better Solution - O(n)

Problem Statement

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.

Brute Force - O(n^2)

public int maxProfit(int[] prices) {
  int l=prices.length;
  int maxProfit = 0;

  for(int i=0; i<l-1; i++) {
    for(int j=i+1; j<l; j++) {
      int diff = prices[j] - prices[i];
      if (diff > maxProfit) {
        maxProfit = diff;
      }
    }
  }
  
  return maxProfit;
}

Better Solution - O(n)

WWe can keep a track of minimum price as we iterate. And, calculate the profit till now. And we can also keep track of maximum profit so far.

The code is pretty simple.

public int maxProfit(int[] prices) {
  int l=prices.length;

  int maxProfit = 0;
  int minPrice = prices[0];
  for(int i=1; i<l; i++) {
    maxProfit = Math.max(maxProfit, prices[i]-minPrice);
    minPrice = Math.min(minPrice, prices[i]);
  }

  return maxProfit;
}