121. Best Time to Buy and Sell Stock

Q: Why else-if and not two separate if statements?

If the price is a new minimum, you can't compute a profit using THIS price as the sell — there's no earlier minimum. The else-if cleanly captures that: update min OR check for new best, not both.

Q: Does it matter where I start minPrice?

Setting it to Infinity (or prices[0]) is safe — the first iteration always sets it correctly. Setting it to 0 would silently allow nonsensical profits on the first day.

easyAsked at Apple

Best Time to Buy and Sell Stock is Apple's canonical single-pass max-tracking question. The trick is recognizing this as 'maximum running-difference' — you only need the minimum-so-far and the best diff. The O(n^2) brute force is mentioned and rejected; the linear scan is the answer.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-19

Source citations

Public interview reports confirming this problem appears in Apple loops.

Glassdoor (2026-Q1)— Apple SWE phone-screen reports list this as the most-repeated array-scan easy.
Blind (2025-12)— Apple new-grad reports cite Best Time to Buy and Sell Stock as the 10-minute warm-up.

Problem

You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock. Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, 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

Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2

Input

prices = [7,6,4,3,1]

Output

Explanation: In this case, no transactions are done and the max profit = 0.

Approaches

1. Brute-force (every pair)

For each buy day, scan all later days to find max profit.

Time: O(n^2)
Space: O(1)

function maxProfit(prices) {
  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: Quadratic; times out on the upper-bound constraints. Mention it and reject.

2. Single pass with running min (optimal)

Track minPrice seen so far and best profit. At each day, update best = max(best, price - minPrice), then update minPrice = min(minPrice, price).

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: Single linear pass, constant space. The else-if structure is correct because if p drops the minPrice, we can't make profit using THIS p as a sell — the next iteration's p might. Apple's preferred form.

Apple-specific tips

Apple grades the single-pass mental model: 'at every position, the best profit ending HERE is price - (the minimum seen so far).' Saying that sentence makes the linear scan inevitable. Then it's just 'track that running min as we go.' Bonus: mention that this generalizes to maximum subarray (Kadane's) if you treat the prices as a delta sequence.

Common mistakes

Allowing prices[j] - prices[i] with j < i (selling before buying).
Initializing best to -Infinity — the problem says return 0 when no profit, so 0 is the right initial value.
Tracking maxPrice instead of minPrice — that solves a different (and trivial) problem.

Follow-up questions

An interviewer at Apple may pivot to one of these next:

Best Time to Buy and Sell Stock II (LC 122) — unlimited transactions.
Best Time to Buy and Sell Stock III (LC 123) — at most 2 transactions, DP.
Best Time to Buy and Sell Stock with Cooldown (LC 309) — state-machine DP.

Solve it now

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Output

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FAQ

Why else-if and not two separate if statements?

If the price is a new minimum, you can't compute a profit using THIS price as the sell — there's no earlier minimum. The else-if cleanly captures that: update min OR check for new best, not both.

Does it matter where I start minPrice?

Setting it to Infinity (or prices[0]) is safe — the first iteration always sets it correctly. Setting it to 0 would silently allow nonsensical profits on the first day.

Free learning resources

Curated free links for this problem.

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