Best Time To Buy And Sell Stock Leetcode

The best time to buy and sell stock leetcode problem explained in depth

The best time to buy and sell stock leetcode problem is a classic algorithmic challenge that asks you to determine the maximum profit achievable by buying a stock on one day and selling it on a later day. Given an array of daily stock prices, you must find the optimal pair of days that yields the highest profit, or return 0 if no profitable transaction exists. This article walks you through the problem’s core concepts, step‑by‑step solution strategies, the underlying mathematical reasoning, common pitfalls, and frequently asked questions, ensuring you can solve it confidently in any coding interview.

Understanding the problem

Problem statement

You are given an integer array prices where each element represents the stock price on a specific day. Your task is to pick one day to buy and a different later day to sell the stock so that the difference prices[sell] - prices[buy] is maximized. If the array is in strictly decreasing order, the maximum profit is 0 because you cannot make a profit The details matter here..

Key constraints

• The array length can be up to 10⁵, so an O(n²) brute‑force approach is infeasible.
• You may only complete one transaction (one buy and one sell).
• Prices are non‑negative integers, but the algorithm works for any numeric type.

Step‑by‑step solution

Below is a concise roadmap that leads to the optimal O(n) solution Worth keeping that in mind..

1. Initialize two variables:

   • min_price to track the lowest price encountered so far.
   • max_profit to store the best profit found.

2. Iterate through the price array once (single pass). For each price p:

   • Update min_price if p is lower than the current min_price.
   • Compute the potential profit p - min_price.
   • If this profit exceeds max_profit, update max_profit.

3. Return max_profit after the loop finishes Turns out it matters..

This greedy, one‑pass method guarantees that you always consider the best buying price up to the current day before evaluating the selling price, which is why it works Simple, but easy to overlook..

Why the one‑pass greedy approach is optimal

• Monotonicity: As you move forward in time, the lowest price seen so far can only decrease or stay the same. So, any future selling price should be compared against this minimal buying price to achieve the greatest possible profit.
• No look‑ahead needed: Since you are allowed only one transaction, the decision to buy at day i depends solely on the minimum price observed up to i. There is no benefit in “waiting” for a better buying opportunity after you have already identified a lower price.

Pseudocode

function maxProfit(prices):
    min_price = +∞
    max_profit = 0
    for p in prices:
        min_price = min(min_price, p)
        profit = p - min_price
        max_profit = max(max_profit, profit)
    return max_profit

Implementation tips

• Edge cases: If the array is empty or has a single element, return 0 immediately.
• Data types: Use a large enough integer type (e.g., long in Java) to avoid overflow when prices are huge.
• Language‑specific nuances: In Python, you can initialize min_price with float('inf'); in JavaScript, use Number.MAX_SAFE_INTEGER.

Common pitfalls

• Resetting min_price incorrectly: Do not reset min_price to the current price after computing profit; you must keep the smallest value seen so far.
• Confusing “sell before buy”: Ensure the sell day index is always greater than the buy day index; the algorithm inherently respects this because min_price is derived from earlier days only.
• Misinterpreting the problem: Some variants allow multiple transactions; the LeetCode version restricts you to a single buy‑sell pair, so do not apply multi‑transaction logic.

Scientific explanation

At its core, the problem models a monotonic sequence where the buying price must be the minimum of a prefix of the array. The proof of optimality relies on the fact that for any day j, the best buying day i (with i < j) is the day with the minimum price among days 0…j‑1. Here's the thing — this is a classic example of the prefix minimum technique, which is widely used in linear‑time scanning algorithms. Because of that, by maintaining the smallest prefix value, you transform a potentially quadratic comparison (checking every pair) into a constant‑time update per element. Hence, the algorithm’s linear scan captures the globally optimal pair without exhaustive enumeration Worth knowing..

Counterintuitive, but true Easy to understand, harder to ignore..

FAQ

Q1: What if the prices are constantly decreasing?
A: The algorithm will keep updating min_price to each new lower value, but profit will always be non‑positive. Since max_profit starts at 0, the final result remains 0, indicating no profitable transaction exists.

Q2: Can I solve this using dynamic programming?
A: Yes, a DP approach with dp[i] representing the maximum profit up to day i works, but it still runs in O(n) time and O(1) space, essentially mirroring the greedy method. The one‑pass greedy is more space‑efficient Simple, but easy to overlook..

Q3: Does the solution handle multiple valid answers?
A: The problem only requires the maximum profit value, not the actual buy and sell days. If you need the indices, keep track of the day when min_price updates and the day that yields max_profit.

Q4: Is there any scenario where a brute‑force solution is acceptable?
A: Only for tiny input sizes (e.g., less than 10 elements) during debugging. For production code or interview settings, the O(n) solution is expected.

Q5: How does this problem relate to real‑world stock trading?
A: It abstracts the idea of finding the most profitable single trade given historical price data, which is a simplified model of actual trading strategies. Real markets involve transaction fees, limits, and multiple trades, making the problem more complex.

Conclusion

Mastering the best time to buy and sell stock leetcode problem equips you with a fundamental algorithmic pattern—maintaining a running minimum while scanning the array once. Plus, this technique not only solves the specific LeetCode challenge efficiently but also forms the backbone of many other optimization problems involving time‑series data. By understanding the underlying monotonic property, implementing the one‑pass greedy approach, and avoiding common mistakes, you can achieve O(n) time complexity with O(1) extra space, a combination that impresses interviewers and performs well in real‑world coding environments. In real terms, remember to test your solution with diverse inputs, handle edge cases, and keep the code clean and readable. With these practices, you’ll be ready to tackle the problem confidently and consistently achieve the highest possible profit.