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9. Best Time to Buy and Sell Stock

easyAsked at Mercury

Find the maximum profit from a single buy-then-sell pair in a price series.

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

Problem

You are given an array prices where prices[i] is the price on day i. Choose one day to buy and a later day to sell to maximize profit; return the maximum profit, or 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

Check every (buy, sell) pair where sell > buy.

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

Keep the running minimum price and update the best profit each step. Linear time and constant space.

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

Tradeoff:

Mercury-specific tips

Mercury reframes this as picking the best FX entry/exit on a treasury wire — defend against negative-profit edge cases the same way you defend against zero-balance edge cases on a startup account.

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Output

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