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121. Best Time to Buy and Sell Stock

easyAsked at JPMorgan

Given a day-indexed price array, find the maximum profit from one buy and one later sell. JPMorgan asks this on nearly every market-data and equities-tech SDE loop because the optimal single-pass min-tracking solution generalises directly to streaming P&L computation.

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

Source citations

Public interview reports confirming this problem appears in JPMorgan loops.

  • Glassdoor (2026-Q1)— Repeated in JPMorgan equities/markets-tech SDE phone-screen and onsite reports through Q1 2026.
  • LeetCode (2026-Q1)— One of the most-tagged JPMorgan problems on the LeetCode company tag page.
  • Blind (2025-12)— Recurring JPMC SWE-II 'classic warm-up' across multiple onsite write-ups.

Problem

You are given an array prices where prices[i] is the price of a given stock on the i-th 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
5

Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6 - 1 = 5.

Example 2

Input
prices = [7,6,4,3,1]
Output
0

Explanation: No transaction is done; max profit = 0.

Approaches

1. Brute-force every pair

Check every (buy_day, sell_day) pair with sell > buy and track the max profit.

Time
O(n^2)
Space
O(1)
function maxProfitBrute(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: Conceptually simplest but TLEs at the 10^5 constraint. Useful only as a warm-up to motivate the single-pass minimum-tracking version.

2. Single pass tracking running minimum (optimal)

Maintain the minimum price seen so far. For each day, candidate profit = today - minSoFar. Take the max.

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

Tradeoff: O(n) time, O(1) space, single forward pass — the production-grade answer for any streaming-price feed. JPMorgan interviewers explicitly note this generalises to running-P&L on tick data.

JPMorgan-specific tips

On markets-tech and equities-tech loops, JPMorgan interviewers ask this and then immediately follow up with 'now imagine prices arrives as a stream — what changes?' The expected answer is 'nothing, the algorithm is already online and O(1) state per tick'. Articulating that connection on the first answer (without waiting for the follow-up) is what separates a strong-hire from a hire on the rubric.

Common mistakes

  • Initialising the running minimum to prices[0] then iterating from index 0 again (double-counts day 0).
  • Updating profit before checking the new minimum, missing the case where today is the new low.
  • Forgetting that the sell must come strictly after the buy (returning a positive value when prices is monotonically decreasing).
  • Returning a negative number instead of 0 when no profit is possible.

Follow-up questions

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

  • What if you can buy and sell multiple times? (LC 122 — sum every positive delta.)
  • What if you can do at most 2 transactions? (LC 123 — two-state DP.)
  • What if you can do at most k transactions? (LC 188 — k-state DP.)
  • What if there is a cooldown day after each sell? (LC 309 — state machine.)
  • What if the prices array arrives as a stream and you must answer max profit so far after every tick?

Solve it now

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Output

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FAQ

Why does the running-minimum approach work in one pass?

The optimal sell day j has profit prices[j] - min(prices[0..j-1]). If you maintain the running min while scanning left-to-right, every j sees the correct min for its prefix, so taking the max over all j gives the global optimum.

Does JPMorgan accept the DP framing (Kadane-style) for this problem?

Yes. Many candidates treat it as 'max subarray of price deltas' which is equivalent. Either framing is accepted, but the running-min explanation tends to land better with markets-tech interviewers because it maps directly to how a trading book actually tracks realised gain.

Free learning resources

Curated free links for this problem.

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