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4. Best Time to Buy and Sell Stock

easyAsked at Rappi

Find the max profit from one buy and one sell in a price series — Rappi frames this as picking the best courier-rate window to lock in for a delivery cohort.

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

Problem

Given an array prices where prices[i] is the price on day i, return the maximum profit you can achieve from buying on one day and selling on a later day. Return 0 if no profit is possible.

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) day pair.

Time
O(n^2)
Space
O(1)
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. Single-pass min tracker

Track the minimum price seen so far and the best profit if you sold today; one sweep suffices.

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

Tradeoff:

Rappi-specific tips

Rappi will press you to defend the single-pass invariant out loud — they want couriers' price-window decisions made in O(n) so the dispatch loop stays under budget.

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Output

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