Best Time to Buy and Sell Stock

To solve this problem, you’ll need to identify the maximum difference between any two numbers in the prices array, where the second number must come after the first. The first number represents the buying price, and the second number represents the selling price.

Here’s a step-by-step guide to finding the maximum profit:

  1. Initialization: Initialize two variables, min_price and max_profit. min_price is used to keep track of the minimum price so far, and max_profit is used to store the maximum profit.
  2. Iteration: Iterate through the prices array from left to right.
    • For each price, calculate the profit by subtracting min_price from the current price.
    • Update max_profit if the calculated profit is greater than the current max_profit.
    • Update min_price if the current price is less than min_price.
  3. Result: max_profit holds the maximum profit that can be achieved.

Here’s the code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # Initialize min_price as the first price and max_profit as 0
        min_price = prices[0]
        max_profit = 0

        # Iterate through the prices from the second price onwards
        for price in prices[1:]:
            # Calculate profit for the current price
            profit = price - min_price

            # Update max_profit if the calculated profit is greater
            max_profit = max(max_profit, profit)

            # Update min_price if the current price is less
            min_price = min(min_price, price)

        return max_profit

Key Takeaways:

  • The code keeps track of the minimum price and calculates the profit for each price.
  • It uses a single loop to iterate through the prices array, so the time complexity is (O(n)), where (n) is the length of the prices array.
  • The space complexity is (O(1)) as we only use a constant amount of extra space.

Abstract Representation of the Problem

An abstract representation of a problem helps in understanding the underlying structure and key elements without getting caught up in specific real-world details. Here’s an abstract representation of the given problem:

  • Elements: A sequence of numbers representing values at discrete time intervals.
  • Objective: Find two distinct indices, (i) and (j), where (i < j), that maximizes the difference between the values at those indices, i.e., maximize the expression (a_j - a_i), where (a) is the given sequence.
  • Constraints: You must choose the indices in the order they appear in the sequence (i.e., you must choose (i) before (j)).

Description

You are provided with a sequence of numbers, where each number represents a value at a specific discrete time interval. Your goal is to select two different points in time, (i) and (j), with (i < j), so that the difference between the values at these two points is maximized.

If there are no such points that result in a positive difference, the result is zero.

This abstract representation focuses on the mathematical and structural aspects of the problem without referring to the real-world concept of buying and selling stocks. It emphasizes the sequence and the objective of maximizing a particular difference under specific constraints.

  THINKING MODE: SYSTEM ANALYST
    1. Identify the invariants
        1. You cannot short sell
        2. Sell price must be higher than the buy price
        3. At most one transaction
        
    2.  [7,1]  --> 0
        [7,10] --> 3
        []     --> 0
  
  THINKING MODE: PROGRAMMER
    1.  Look for min price that I can buy the stock
            buy price is the min 
            update the min if we encounter a new min price
            
    2.  Keep track of max profit as we traverse through the array
            4
            2
            5
            3

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19

# @param {Integer[]} prices
# @return {Integer}

def max_profit(prices)
    if prices.size < 2
        return 0
    end

    min = prices[0]
    profit = 0

    for i in (1..prices.size-1)
        new_profit = (prices[i] - min)
        profit = [new_profit, profit].max
        min = [min, prices[i]].min
    end

    profit
end

Identifying Problem Isomorphism

“Best Time to Buy and Sell Stock” can be isomorphically mapped to “Maximum Subarray”.

Reasoning: Both problems are asking for a way to maximize a certain outcome based on a series of choices.

In “Best Time to Buy and Sell Stock”, we’re given a series of prices and we need to decide when to buy and when to sell to maximize profit. The solution involves finding a minimum price (buying point) and a maximum price (selling point) after the buying point.

In “Maximum Subarray”, we’re given a series of numbers, and we’re asked to find a contiguous subarray that has the greatest sum. This is similar to finding the lowest buying point and the highest selling point in the “Best Time to Buy and Sell Stock” problem.

The mapping is as follows: a stock’s price on day i maps to the i-th number in the array in “Maximum Subarray”, the profit from buying and selling the stock maps to the maximum subarray sum.

This mapping is approximate since in “Maximum Subarray” we allow subarrays of length 1 (i.e., buying and selling on the same day), while in “Best Time to Buy and Sell Stock” we’re asked to return 0 profit if no transaction is done (i.e., all prices are in decreasing order). But generally, the “Maximum Subarray” problem is a good isomorphic mapping due to the similarity in finding the maximum profit/sum in a series of numbers.

“Best Time to Buy and Sell Stock” is a problem that involves dynamic programming and array manipulation. Here are 10 problems that can help you prepare for it:

  1. 53. Maximum Subarray: This problem involves finding a subarray with the maximum sum, which is similar to finding a period with the highest profit in the stock problem.

  2. 121. Best Time to Buy and Sell Stock II: This problem is a slight variation of the original one, allowing multiple transactions.

  3. 122. Best Time to Buy and Sell Stock III: This problem adds a new constraint to the original problem - maximum of two transactions.

  4. 70. Climbing Stairs: This problem involves dynamic programming to find the number of ways to reach a certain point, which can be helpful for understanding dynamic programming concepts.

  5. 198. House Robber: This problem also involves dynamic programming to find the maximum sum of non-adjacent numbers.

  6. 628. Maximum Product of Three Numbers: This problem can help you practice array manipulation and finding maximum/minimum values.

  7. 674. Longest Continuous Increasing Subsequence: This problem involves finding an increasing subsequence in an array, which is similar to finding a period of time to buy and sell stocks for maximum profit.

  8. 746. Min Cost Climbing Stairs: This problem involves dynamic programming with the objective to minimize a cost, which can help you understand how to approach optimization problems.

  9. 300. Longest Increasing Subsequence: This problem requires finding the longest increasing subsequence, which is a more complex version of finding an increasing subsequence.

  10. 509. Fibonacci Number: This problem involves a simple dynamic programming scenario and can be a good starting point for understanding dynamic programming.

These cover dynamic programming, array manipulation, and maximum/minimum subarray problems, which are useful for “Best Time to Buy and Sell Stock”.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

def maxProfit(self, prices: List[int]) -> int:
	if not prices:
		return 0

	maxProfit = 0
	minPurchase = prices[0]

	for i in range(1, len(prices)):		
		maxProfit = max(maxProfit, prices[i] - minPurchase)
		minPurchase = min(minPurchase, prices[i])
	return maxProfit

Problem Classification

This problem falls within the “Financial Analysis” domain, as it deals with stock prices and the idea of buying and selling to maximize profit. It is a “maximization” problem under the broad category of “Optimization Problems”.

Here are the ‘What’ components of the problem:

  1. Input: The input is an array of stock prices. Each element in the array represents the stock price for a given day.

  2. Output: The output is a single integer that represents the maximum profit one can achieve by buying and selling the stock.

  3. Constraints: The number of elements in the prices array is between 1 and 10^5 (inclusive), and each element of prices is a non-negative integer up to 10^4.

We can further classify the problem as follows:

  • It is a “Single-Answer” problem because the goal is to find a single correct answer (maximum profit).
  • It is a “Sequential” problem because the order of the stock prices matters, i.e., buying has to be done before selling.
  • It is a “Non-constructive” problem, as we don’t have to produce a specific method or sequence of buying and selling, but rather find the maximum possible profit.
  • This problem could also be considered as a “Dynamic Programming” problem because it involves finding an optimal solution by breaking down the problem into simpler, overlapping sub-problems (though this relates more to the ‘how’ of solving it).
  • It is not an “Adversarial” problem as it does not involve any competition or opponent.

The task is to analyze the given array and find the maximum difference between any two elements, with the condition that the smaller element’s index is less than the larger element’s index. Thus, it’s also an “Array Processing” problem.

This problem can be classified based on the problem domain as follows:

  1. Dynamic Programming: This problem involves finding an optimal solution by making decisions at each step based on the best choice at that point in time, which is a characteristic of dynamic programming problems. Here, we decide whether to sell the stock or keep it based on whether it would result in maximum profit or not.

  2. Array Processing: The problem involves processing an array (or list) of stock prices. This requires understanding of array traversal and manipulation.

  3. Optimization Problem: At its core, this problem is about optimizing a quantity - in this case, the profit from stock transactions. It involves finding the maximum profit that can be obtained from the given list of prices, which is an optimization task.

Clarification Questions

What are the clarification questions we can ask about this problem?

Visual Model of the Problem

To visualize the “Best Time to Buy and Sell Stock” problem, you can create a line graph with time on the x-axis and price on the y-axis.

Consider an array for the following 7 days: [7,1,5,3,6,4]

  1. Start by marking each point on the graph:

    Day 1 (x=1, y=7) Day 2 (x=2, y=1) Day 3 (x=3, y=5) Day 4 (x=4, y=3) Day 5 (x=5, y=6) Day 6 (x=6, y=4)

  2. Connect these points to form the line graph.

The problem now becomes finding the pair of points that form the steepest uphill line segment (which represents the largest increase in stock price).

In this case, the pair is (Day 2, Day 5) as buying at price 1 and selling at price 6 yields the maximum profit of 5. This corresponds to the steepest uphill line segment on the graph.

Constraints

Given the constraints and the problem statement, there are several characteristics that we can use to our advantage to find an efficient solution:

  1. Sequential Order: The fact that the prices are given in a sequential order can be exploited. Since the buying date should be before the selling date, we need to always look forward from a given position in the array, not backwards. This naturally leads us to iterative solutions which have a time complexity of O(n), where n is the size of the input array.

  2. Range of Prices: Each price in the array is a non-negative integer up to 10^4. This is useful information, but it doesn’t immediately suggest any particular computational shortcut or data representation trick.

  3. Array Size: The array length is between 1 and 10^5. This tells us that the solution must be quite efficient, ruling out algorithms with quadratic time complexity or worse for large inputs.

  4. Maximum and Minimum Values: One key insight here is that to maximize profit, one must buy at the minimum price and sell at the maximum price. However, the maximum price must occur after the minimum price. Hence, we can keep track of the minimum price encountered so far and the maximum profit we can get at each day.

  5. Non-Negative Prices: Since prices are non-negative, the minimum possible price is 0. This fact can be used in initializations when we’re trying to find the minimum price so far.

  6. Single Transaction: The problem states we can complete at most one transaction. This prevents us from using strategies that involve multiple buying and selling operations.

These insights can help design an efficient algorithm to solve the problem.

Thought Process

Let’s look at the expensive that breaks down a dynamic programming solution. Here, we’re only allowed to make one transaction - we must sell the stock after we’ve bought it.

This problem might initially seem suited to a brute-force solution, where you calculate the profit for every possible pair of buy-sell days using two nested loops. However, this is unnecessary and computationally expensive.

The solution described applies a technique known as the Sliding Window, a common approach in dynamic programming problems where we consider a sub-array or sublist and slide it along our given list to find an optimal solution.

In the context of this problem, as we ‘slide’ through the list of prices, we need to consider two aspects:

  1. Is the current price the lowest we’ve seen so far? If it is, this would be the best day to buy the stock.
  2. If we sold the stock today (at the current price), would we make a bigger profit than any profit we’ve seen so far? This profit is calculated as the difference between the current price and the lowest price we’ve seen.

An important note is that if condition 1 is true (i.e., if today’s price is the lowest we’ve seen), condition 2 cannot also be true, because if we bought the stock today, we couldn’t also sell it today.

Let’s apply this logic to a sample list of prices: [4,1,5,2,7]

  • On the first day, the price is 4. This is the lowest price we’ve seen, but since we can’t sell on the same day we buy, the maximum profit is still 0.
  • On the second day, the price is 1. This is now the lowest price we’ve seen, but if we sold today, we’d make a loss, so the maximum profit remains 0.
  • On the third day, the price is 5. This isn’t lower than 1, but if we sold today (having bought the stock when the price was 1), we’d make a profit of 4. This is the highest profit we’ve seen so far.
  • On the fourth day, the price is 2. This isn’t lower than 1, and if we sold today (having bought when the price was 1), we’d make a profit of only 1, so there’s no need to update our maximum profit.
  • On the fifth day, the price is 7. This isn’t lower than 1, but if we sold today (having bought when the price was 1), we’d make a profit of 6, so we update our maximum profit to 6.

So, at the end of this process, the maximum profit we could make from one buy-sell transaction is 6.

The problem statement is asking us to calculate the maximum profit we can achieve by buying a stock at a certain day and selling it on any day after that. In other words, we need to find a pair of indices i and j (i < j) such that the difference prices[j] - prices[i] is maximized.

This can be interpreted as a difference of two elements in an array problem where the maximum difference between two elements is required, but the larger element must appear after the smaller one.

We start by initializing min_price as the first element of the array. This represents the lowest buying price we have encountered so far. We also initialize max_profit to be 0, as we haven’t calculated any profit yet.

We then iterate through the array. For each element:

  1. We first update min_price if the current element is lower. This means we have found a potentially better day to buy the stock.
  2. We then calculate the profit if we were to sell the stock at the current price, and update max_profit if this profit is greater than the current max_profit.

By iterating through the array in this way, we ensure that we always calculate profits for selling prices that occur after the buying price, as required by the problem statement.

At the end of this process, max_profit will contain the maximum profit that can be achieved.

Python Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

def maxProfit(prices):
    min_price = prices[0]
    max_profit = 0
    
    for i in range(1, len(prices)):
        # If current price is less than min_price, update min_price
        if prices[i] < min_price:
            min_price = prices[i]

        # If profit from selling at current price is more than max_profit, update max_profit
        elif prices[i] - min_price > max_profit:
            max_profit = prices[i] - min_price

    return max_profit

Insights about the Problem Statement

The problem is a classic example of dynamic programming, specifically a maximization problem. The cue in the problem statement is the requirement to maximize profit, which is a common theme in optimization problems. This, along with the constraint that the buying day must be before the selling day, indicates that a dynamic programming approach would be suitable. The problem statement also indirectly indicates that there can be days when no transactions may be done, implying that the maximum profit could be 0. This should be kept in mind when designing the solution.

Solution Approach and Analysis

Given the problem statement, the goal is to find the maximum difference between two elements in the array such that the smaller element appears before the larger element, because the smaller element corresponds to the buying price and the larger one corresponds to the selling price. Here are the steps we can use to find the solution:

  1. Initialization: Initialize two variables, min_price and max_profit. min_price is set to the highest possible value (float('inf') or prices[0]) and max_profit is set to 0. min_price keeps track of the minimum price seen so far and max_profit stores the maximum profit we can get.

  2. Iteration: Iterate over the prices array from left to right. For each price, we do two things:

    a. Update min_price: If the current price is less than min_price, then we update min_price with the current price.

    b. Update max_profit: If the difference between the current price and min_price is greater than max_profit, we update max_profit.

  3. Result: After iterating through all prices, max_profit will hold the maximum profit achievable from one transaction. If no profitable transaction is possible, max_profit will remain 0.

Explanation

Think of it like you’re a trader watching the stock market. min_price is the lowest price you’ve seen so far, and you consider buying at that price. When you see a new price, you think, “If I sell at this price, how much profit will I make?” (current price - min_price). If this profit is more than any profit you’ve calculated before (max_profit), then you update your selling plan.

If the new price is lower than your potential buying price (min_price), you’ll consider changing your plan and buying at this lower price instead.

Effect of Changes in Parameters

Changes in the price values or the order of the prices can significantly impact the maximum profit. However, the size of the input (prices.length) mainly affects the time it takes to find the solution, not the solution itself.

Working Example

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

  • Start with min_price = 7 and max_profit = 0
  • The 2nd price is 1, which is less than min_price, so min_price becomes 1.
  • The 3rd price is 5. The profit from selling at 5 when bought at 1 (5-1=4) is greater than max_profit (0), so max_profit becomes 4.
  • The 4th price is 3, but it doesn’t change min_price or max_profit.
  • The 5th price is 6. The profit from selling at 6 when bought at 1 (6-1=5) is greater than max_profit (4), so max_profit becomes 5.
  • The last price is 4, but it doesn’t change min_price or max_profit.
  • Therefore, the maximum profit is 5.

Language Agnostic Drills and Targeted Drills in Python

Drill 1 - Understanding Lists: Practice traversing through a list, accessing elements, and getting familiar with list indices.

1
2
3

def print_elements(prices):
    for price in prices:
        print(price)

Drill 2 - Using if conditionals: Understand the usage of if conditions for logical operations.

1
2
3
4
5

def check_price(price):
    if price > 100:
        print("Expensive")
    else:
        print("Cheap")

Drill 3 - Finding the minimum value in a list: Develop a function that can find the minimum value in a list of numbers.

1
2
3
4
5

def find_minimum(prices):
    min_price = prices[0]
    for price in prices:
        min_price = min(min_price, price)
    return min_price

Drill 4 - Finding the maximum value in a list: Similar to the previous drill, develop a function that can find the maximum value in a list.

1
2
3
4
5

def find_maximum(prices):
    max_price = prices[0]
    for price in prices:
        max_price = max(max_price, price)
    return max_price

Drill 5 - Calculating Profit: Create a function that calculates the profit given a list of buy and sell prices.

1
2
3
4
5
6
7
8

def calculate_profit(prices):
    min_price = prices[0]
    max_profit = 0
    for price in prices:
        min_price = min(min_price, price)
        potential_profit = price - min_price
        max_profit = max(max_profit, potential_profit)
    return max_profit

The last drill calculate_profit is your final solution. Each drill focuses on a particular coding concept and progressively builds up to the final solution.