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14. Best Time to Buy and Sell Stock

easyAsked at Figma

Find the maximum profit from a single buy-and-sell on a price array. Figma uses it as a tempo-setter to check whether you collapse two nested loops into a running minimum.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

You are given an array prices where prices[i] is the price of a stock on day i. Return the maximum profit you can achieve from buying on one day and selling on a later day. If no profit is possible, return 0.

Constraints

  • 1 <= prices.length <= 10^5
  • 0 <= prices[i] <= 10^4

Examples

Example 1

Input
prices = [7,1,5,3,6,4]
Output
5

Example 2

Input
prices = [7,6,4,3,1]
Output
0

Approaches

1. Brute force pairs

Try every (buy, sell) pair and keep the max difference.

Time
O(n^2)
Space
O(1)
function maxProfit(prices) {
  let best = 0;
  for (let i = 0; i < prices.length; i++)
    for (let j = i + 1; j < prices.length; j++)
      best = Math.max(best, prices[j] - prices[i]);
  return best;
}

Tradeoff:

2. Running minimum

Walk once, tracking the smallest price seen so far and the best profit achievable selling today. One pass, constant space.

Time
O(n)
Space
O(1)
function maxProfit(prices) {
  let minPrice = Infinity;
  let best = 0;
  for (const p of prices) {
    if (p < minPrice) minPrice = p;
    else if (p - minPrice > best) best = p - minPrice;
  }
  return best;
}

Tradeoff:

Figma-specific tips

Figma values verbal narration of the invariant ('best buy seen so far is monotone non-increasing') — say it out loud before you write the loop body.

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Output

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