This Python code is a solution to the well-known "Best...

This Python code is a solution to the well-known "Best Time to Buy and Sell Stock" problem (often presented in coding interviews such as on LeetCode).

What It Does:

The code calculates the maximum profit you can achieve by buying and selling a stock, given an array of stock prices (prices) where the ( i )-th element represents the stock price on day ( i ). You are only allowed to make one buy and one sell (no multiple transactions or shorting).

Explanation of the Code:

1. Variables:

   • min_price: Keeps track of the lowest stock price seen so far as you iterate through the prices. It starts as infinity (float('inf')), because initially we haven't seen any price.
   • max_profit: Stores the maximum profit possible so far. It starts at 0 because the maximum profit at the beginning is zero.

2. Logic (Loop through prices):

   • For each price in the prices array:
     1. Check for a new minimum price: If the current price is less than min_price, update min_price to be this price. This simulates buying the stock at the lowest price seen.
     2. Update the maximum profit: If the current price is greater than the min_price, calculate the potential profit (price - min_price) and update max_profit if this profit is larger than the value currently stored in max_profit.

3. Result:

   • After the loop, max_profit contains the highest profit possible from one buy-sell transaction.

Example Walkthrough:

Input:

prices = [7, 1, 5, 3, 6, 4]

• Day 1: price = 7
  • min_price = 7, max_profit = 0 (no profit yet)
• Day 2: price = 1
  • min_price = 1 (new minimum), max_profit = 0
• Day 3: price = 5
  • Profit = ( 5 - 1 = 4 ), update max_profit to 4
• Day 4: price = 3
  • Profit = ( 3 - 1 = 2 ), max_profit remains 4
• Day 5: price = 6
  • Profit = ( 6 - 1 = 5 ), update max_profit to 5
• Day 6: price = 4
  • Profit = ( 4 - 1 = 3 ), max_profit remains 5

Output:

max_profit = 5

Edge Cases:

1. Empty Prices (prices = []):

   • The loop does not execute, so the function returns max_profit = 0.

2. One Price (prices = [5]):

   • No transactions are possible, so the profit is 0.

3. Prices Always Decreasing (prices = [7, 6, 5, 4]):

   • The minimum price keeps decreasing, but no profitable transaction can be made, so profit remains 0.

Overall Complexity:

• Time Complexity: ( O(n) ), where ( n ) is the length of the prices array (only one pass is made through the array).
• Space Complexity: ( O(1) ), as only a few variables are used (no additional data structures).