Best Time To Buy And Sell Stock II Leetcode Problem 122 [Python]
Welcome to another coding adventure!
In this blog post, we will tackle the Best Time To Buy And Sell Stock II problem, which is a LeetCode problem with the identifier 122. We will explore the problem, break it down, and provide you with an efficient Python solution.
By the end of this post, you'll have a clear understanding of how to approach and solve this problem.
Problem Overview
The problem statement is as follows:
**You are given an integer array prices where prices[i] is the price of a given stock on the i-th day.
On each day, you may decide to buy and/or sell the stock.
You can only hold at most one share of the stock at any time.
However, you can buy it then immediately sell it on the same day.
Find and return the maximum profit you can achieve.**
In other words, you're given a list of stock prices, and your goal is to maximize your profit by buying and selling the stock optimally.
Example 1
Let's consider the first example to understand the problem better.
Input:
prices = [7, 1, 5, 3, 6, 4]
Output:
7
Explanation:
You can buy on day 2 (price = 1) and sell on day 3 (price = 5), which results in a profit of 5 – 1 = 4. Then, you can buy on day 4 (price = 3) and sell on day 5 (price = 6), resulting in a profit of 6 – 3 = 3. The total profit is 4 + 3 = 7. This example demonstrates that you can buy and sell the stock multiple times to maximize your profit.
Understanding Constraints
Before diving into the solution, let's break down the constraints provided in the problem statement:
1 <= prices.length <= 3 * 10^4: This tells us that the input list of stock prices can have a length ranging from 1 to 30,000.0 <= prices[i] <= 10^4: The prices of the stock on each day are non-negative integers, with values ranging from 0 to 10,000.
Best Time to Buy And Sell Stock II LeetCode Problem Solution
Now that we understand the problem, let's move on to solving it efficiently.
We will start with a brute-force approach and then introduce a more optimized solution.
1. Bruteforce Approach
In the brute-force approach, we would consider all possible transactions and calculate the profit for each.
The maximum profit among all these possibilities will be our answer.
We'll use two nested loops to iterate through the prices list.
The outer loop represents the buy day, and the inner loop represents the sell day.
We'll calculate the profit for each combination and keep track of the maximum profit.
def maxProfit(prices):
max_profit = 0
for buy_day in range(len(prices)):
for sell_day in range(buy_day + 1, len(prices)):
profit = prices[sell_day] - prices[buy_day]
if profit > 0:
max_profit += profit
return max_profit
While this approach works, it is not the most efficient solution, and it has a time complexity of O(n^2), where n is the number of days.
2. An Efficient Approach
Now, let's introduce an optimized approach that has a time complexity of O(n), making it much more efficient.
We'll leverage the fact that we can add the profit for each price increase without performing individual buy and sell operations.
def maxProfit(prices):
max_profit = 0
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
max_profit += prices[i] - prices[i - 1]
return max_profit
In this efficient solution, we iterate through the prices list once, and whenever we find a day where the stock price is higher than the previous day, we add the price difference to our max_profit.
This approach works because we are essentially buying low and selling high, and we accumulate all such profits.
Python Code Solution
Here's the Python code that implements the efficient solution for the Best Time To Buy And Sell Stock II problem:
def maxProfit(prices):
max_profit = 0
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
max_profit += prices[i] - prices[i - 1]
return max_profit
You can use this function to find the maximum profit you can achieve from the given stock prices.
Time and Space Complexity
Let's briefly discuss the time and space complexity of our efficient solution.
- Time Complexity: The time complexity is
O(n), where n is the number of days (length of thepriceslist).
We iterate through the list once, making a constant-time comparison for each day.
- Space Complexity: The space complexity is
O(1)because we only use a constant amount of extra space to store themax_profitvariable and the loop variablei.
Reasoning Behind Our Approach
The reasoning behind our approach is quite simple.
We observe that the goal is to maximize profit, and to do that, we need to identify price increases and accumulate the profits from those increases.
The efficient solution achieves this by iterating through the list of prices and adding the price differences whenever a profitable transaction can be made.
This approach is highly efficient and takes advantage of the simple nature of the problem.
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Conclusion
In this blog post, we tackled the Best Time To Buy And Sell Stock II problem, which is a LeetCode problem with the identifier 122. We started by understanding the problem statement and its constraints.
Then, we provided a brute-force approach to solving the problem, followed by a more efficient solution.
Our efficient solution has a time complexity of O(n), making it a practical and effective way to maximize profits from stock trading.
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Happy coding!
Happy coding, and may your stock trading adventures be profitable!
