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9. Best Time to Buy and Sell Stock

easyAsked at Grab

Find the max profit from a single buy and single sell — Grab uses this as a warm-up for one-pass scanning.

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

Problem

Given an array prices where prices[i] is the price of a given stock on day i, maximize profit by choosing a single day to buy and a future day to 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

Check all pairs (i, j) with i < j.

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. One pass min-tracker

Keep the running minimum buy price; update best profit at each day.

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:

Grab-specific tips

Grab interviewers grade fast on whether you spot the streaming/one-pass pattern — frame the array as ride-fare history over time.

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Output

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