Best Time to Buy and Sell Stock Algorithm

You get one shot. Buy once, sell once. Find the maximum profit.

The Problem

Given an array prices where prices[i] is the stock price on day i, find the best single transaction. You must buy before you sell. If no profit is possible, return 0.

Input: prices = [7,1,5,3,6,4]
Output: 5
Buy at 1 (day 2), sell at 6 (day 5). Profit = 5.

The Brute Force Trap

The naive approach checks every pair: for each buy day, look at every future sell day. That's O(n²). It works, but it's slow.

The Insight

You don't need to check every pair. As you scan left to right, track two things:

1. The lowest price seen so far (your best buy opportunity).
2. The maximum profit if you sold today.

One pass. Done.

The Code

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

  for (let i = 0; i < prices.length; i++) {
    if (prices[i] < minPrice) {
      minPrice = prices[i];
    } else {
      maxProfit = Math.max(maxProfit, prices[i] - minPrice);
    }
  }

  return maxProfit;
}

console.log(maxProfit([7, 1, 5, 3, 6, 4])); // 5
console.log(maxProfit([7, 6, 4, 3, 1]));    // 0

Complexity

• Time: O(n) — single pass.
• Space: O(1) — two variables.

Why It Works

At each price, you're asking: "If I sold right now, what's the best I could have done?" The answer is always currentPrice - lowestPriceSoFar. By tracking the running minimum, you never miss the optimal buy point.

The Trade-off

This only handles one transaction. If you're allowed multiple buys and sells, you need a different approach (see the Stock II problem). If there's a transaction fee or cooldown, you need dynamic programming.