Best Time to Buy and Sell Stock – Coding Interview Problem

You are given an array prices where prices[i] is the price of a given stock on the i-th 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:

Input: [7,1,5,3,6,4]
Output: 5 (Buy on day 2 (price=1) and sell on day 5 (price=6), profit = 6-1 = 5)

Brute-Force Approach

The basic idea is simple: we will use two nested loops to go through each pair of days and calculate the difference between selling and buying prices.

We will update our maximum profit whenever we find a pair that gives us higher profit.

Java Code:

public class StockProfit {
  public static int maxProfit(int[] prices) {
    int maxProfit = 0;
    
    for (int i = 0; i < prices.length; i++) {
      for (int j = i + 1; j < prices.length; j++) {
        int profit = prices[j] - prices[i];
        
        if (profit > maxProfit) {
          maxProfit = profit;
        }
      }
    }
    
    return maxProfit;
  }

  public static void main(String[] args) {
    int[] prices = {7, 1, 5, 3, 6, 4};
    int result = maxProfit(prices);
    System.out.println("Maximum Profit: " + result);  // Output will be 5
  }
}

Complexity Analysis

• Time Complexity: O(n²) Because of two nested for loops.
• Space Complexity: O(1) No extra space used.

Optimal Approach

The optimal approach uses a single loop.

We keep track of the minimum price we’ve seen so far and the maximum profit we can make for each day, with that minimum price.

Java Code:

public class StockProfit {
  public static int maxProfit(int[] prices) {
    int minPrice = Integer.MAX_VALUE;
    int maxProfit = 0;
    
    for (int i = 0; i < prices.length; i++) {
      if (prices[i] < minPrice) {
        minPrice = prices[i];
      } else if (prices[i] - minPrice > maxProfit) {
        maxProfit = prices[i] - minPrice;
      }
    }
    
    return maxProfit;
  }

  public static void main(String[] args) {
    int[] prices = {7, 1, 5, 3, 6, 4};
    int result = maxProfit(prices);
    System.out.println("Maximum Profit: " + result);  // Output will be 5
  }
}

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

Summary

The brute-force approach works but is slow O(n²), making it impractical for large inputs.

The optimal approach gives us the same result much more efficiently O(n), and it only uses constant extra space.