Overview

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 and a different day in the future to sell. Return the maximum profit you can achieve. If no profit is possible, return 0.

Implement maxProfit(prices).

Constraints

• 1 ≤ prices.length ≤ 100000
• 0 ≤ prices[i] ≤ 10000
• You must buy before you sell (buy day < sell day).
• Only one transaction allowed.
• Aim for $O(n)$ time, $O(1)$ space.

Examples

maxProfit([7, 1, 5, 3, 6, 4]);
// → 5  — buy on day 1 (price 1), sell on day 4 (price 6) → 6 - 1 = 5

maxProfit([7, 6, 4, 3, 1]);
// → 0  — prices only decrease, no profitable transaction

maxProfit([2, 4, 1]);
// → 2  — buy day 0, sell day 1

Notes

• Track the minimum price seen so far. At each day, compute the profit if you sold today (price - minSoFar) and track the global max.
• Single pass, constant space.

Solution

function maxProfit(prices) {
  let minPrice = Infinity;
  let maxProfit = 0;

  for (const price of prices) {
    if (price < minPrice) {
      minPrice = price;
    } else {
      maxProfit = Math.max(maxProfit, price - minPrice);
    }
  }

  return maxProfit;
}

Resources

• Best Time to Buy and Sell Stock — LeetCode