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9. Best Time to Buy and Sell Stock

easyAsked at Coupang

Find the max profit from one buy-sell pair, mirroring how Coupang detects best price gaps in fluctuating SKU pricing for peak-event throughput.

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 from a single buy followed by a sell. 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

Try every (buy, sell) 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. Track running min

Sweep once, keeping the minimum price seen so far and the max profit relative to it.

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

Tradeoff:

Coupang-specific tips

Coupang's pricing engine runs single-pass peak detection across SKU price streams during peak-event throughput; one-pass solutions show you understand streaming over batch.

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Output

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