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2. Best Time to Buy and Sell Stock

easyAsked at Confluent

Find the maximum profit from a single buy-sell over an array of daily prices — Confluent maps it to a single-pass min-tracker that mirrors how a Kafka consumer holds running aggregates per partition.

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

Problem

Given an array prices where prices[i] is the price of a stock on day i, return the maximum profit from buying on one day 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

Try every (buy, sell) pair and track max profit.

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 tracker

Walk left-to-right keeping the running minimum price seen; at each step, the best sell-today profit is price minus min.

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:

Confluent-specific tips

Confluent likes when you frame this as a per-partition running aggregate — call out that exactly-once semantics require the min state be checkpointed before emitting the profit downstream.

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Output

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