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121. Best Time to Buy and Sell Stock

easyAsked at Bloomberg

Given a daily price series, return the maximum profit from one buy + one sell. Bloomberg leans on this for finance-engineering candidates — it's a single-pass greedy that mirrors the real-world 'running max profit' calculation traders care about.

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

Source citations

Public interview reports confirming this problem appears in Bloomberg loops.

  • Glassdoor (2026-Q1)— Bloomberg SWE phone-screen reports list this as a near-default question for fintech-team rounds.
  • Blind (2025-11)— Bloomberg new-grad reports highlight it as common for finance-adjacent teams.

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
5

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

Example 2

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

Explanation: No profitable transactions; return 0.

Approaches

1. Single-pass running minimum (optimal)

Track the minimum price seen so far. On each day, profit = price - min. Return the max profit ever seen.

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

Tradeoff: The canonical answer. Single pass, constant space, and the structure mirrors how traders mentally track 'best buy point so far'.

2. Brute-force two-pointer

For every buy day, scan all future sell days. Return the max diff.

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++) {
      if (prices[j] - prices[i] > best) best = prices[j] - prices[i];
    }
  }
  return best;
}

Tradeoff: Easy to reason about. Bloomberg interviewers want you to NAME this first, then show the O(n) version.

Bloomberg-specific tips

Bloomberg interviewers grade this on the SINGLE-PASS INTUITION. State 'on each day I track the cheapest buy point so far' before coding. Avoid Kadane's-style DP framing — for the one-transaction case the running-min approach is more direct.

Common mistakes

  • Returning a negative profit when no profit is possible (return 0 instead).
  • Using max(prices) - min(prices) — that fails when the max comes before the min.
  • Over-engineering with DP when O(1) state suffices.

Follow-up questions

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

  • Best Time to Buy and Sell Stock II (LC 122) — unlimited transactions, greedy.
  • Best Time to Buy and Sell Stock III (LC 123) — at most 2 transactions, DP.
  • Best Time to Buy and Sell Stock IV (LC 188) — at most k transactions.

Solve it now

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Output

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FAQ

Why running min, not running max?

Because each day's potential profit is current_price - cheapest_buy_so_far. You always want the cheapest buy point that comes BEFORE today.

Will Bloomberg ask the multi-transaction follow-up?

Often, yes. Have LC 122 (unlimited) ready — it's a greedy where you sum every up-step.

Free learning resources

Curated free links for this problem.

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