Best Time to Buy and Sell Stock

1. Define the problem

You are given an array prices where prices_i is the price of a given stock on the ith 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. Track the minimum price seen so far in one pass, and compare each day against that running minimum to find the best profit.

Constraints

1 ≤ prices.length ≤ 10⁵
0 ≤ prices_i ≤ 10⁴

Example

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

Output5

Explanation Buy on day 1 (price = 1) and sell on day 4 (price = 6), profit = 6 - 1 = 5.

2. Visualize the solution

Track the running minimum price

Statusinit

Day 0: price 7 — set as running minimum. Best profit = 0.

What happens in this step

minPrice = prices[0] = 7
bestProfit = 0

No later day has been compared yet, so profit starts at 0.

Step 1 of 6

Steps to visualize

Start with the running minimum at the first day's price and best profit at 0.
Move right one day at a time.
If today's price is lower than the running minimum, update the minimum.
Otherwise compute profit as price minus the running minimum and update best if it improves.
Stop after scanning the last day.

3. Walk through the code

Walk through the code

Same walkthrough, now with the code. Press Next to move one step and watch which lines run.

Track the running minimum price

Statusinit

Day 0: price 7 — set as running minimum. Best profit = 0.

What happens in this step

minPrice = prices[0] = 7
bestProfit = 0

No later day has been compared yet, so profit starts at 0.

Step 1 of 6

4. Solution

Solution

solution.tsTypeScript

function maxProfit(prices) {
  if (prices.length === 0) {
    throw new Error("prices must contain at least one price");
  }

  let minPrice = prices[0];
  let bestProfit = 0;

  for (let i = 1; i < prices.length; i++) {
    if (prices[i] < minPrice) {
      minPrice = prices[i];
    } else {
      bestProfit = Math.max(bestProfit, prices[i] - minPrice);
    }
  }

  return bestProfit;
}

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

5. Test cases

Test cases

Input	Expected	Covers
`prices = [7, 1, 5, 3, 6, 4]`	`5`	example from the docstring
`prices = [5]`	`0`	smallest valid input: single day, no transaction possible
`prices = [3, 3, 3, 3]`	`0`	all identical prices
`prices = [9, 7, 5, 3, 1]`	`0`	prices only decrease, so no profit is possible
`prices = [1, 2, 3, 4, 5]`	`4`	prices only increase: best is buy first, sell last
`prices = [0, 0, 0, 0]`	`0`	boundary value: prices can be 0
`prices = [3, 2, 6, 5, 0, 3, 9, 4]`	`9`	larger, hand-verified case

Best Time to Buy and Sell Stock