121. Best Time to Buy and Sell Stock

At a Glance

• Topic: Array
• Pattern: Analyze Pattern
• Difficulty: Easy
• LeetCode: 121

Problem Statement

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.

Example 1:

Input: prices = [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

1 <= prices.length <= 105
0 <= prices[i] <= 104

Approach & Solution Steps

Maintain a variable for the minimum price seen so far. Iterate through the array, updating the minimum price and calculating the maximum profit at each step.

Optimal JS Solution

function maxProfit(prices) {
  let minPrice = Infinity;
  let maxProfit = 0;
  for (const price of prices) {
    if (price < minPrice) minPrice = price;
    else if (price - minPrice > maxProfit) maxProfit = price - minPrice;
  }
  return maxProfit;
}

Edge Cases & Pitfalls

• Always consider empty or null inputs.
• Watch out for off-by-one index errors.