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6. Best Time to Buy and Sell Stock

easyAsked at Swiggy

Find the max profit from one buy-sell of a stock given daily prices.

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

Problem

Given prices[i] = price of a stock on day i, return the maximum profit you can make by buying once 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.

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

Tradeoff:

2. Single pass min tracking

Track running minimum and best profit-if-sold-today as you scan once. Profit = current price minus best buy seen so far.

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:

Swiggy-specific tips

Swiggy maps this onto surge-pricing decisions; the bonus signal is verbalizing why local minima beat tracking pair-wise diffs.

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Output

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