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11. Best Time to Buy and Sell Stock

easyAsked at Intuit

Given an array of stock prices by day, find the maximum profit from a single buy/sell transaction. Intuit asks this to test single-pass min/max tracking and to probe whether you handle financial values as integer cents instead of floats.

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

Source citations

Public interview reports confirming this problem appears in Intuit loops.

  • Glassdoor (2026-Q1)— Intuit Mint team phone screen, framed around portfolio P/L calculations.
  • LeetCode Discuss (2025-10)— QuickBooks SWE intern screen cited Buy/Sell Stock as warm-up.

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
  • Prices are integers representing cents to avoid float drift.

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: Prices only fall; no profitable transaction exists.

Approaches

1. Brute force double loop

Try every (buy, sell) pair where sell > buy and track the maximum difference.

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 — fine to mention but propose the single-pass version immediately.

2. Single-pass min tracking (optimal)

Walk left-to-right, tracking the minimum price seen so far. For each day, the best sell-today profit is price - minSoFar. Keep the running max.

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: Linear time, constant space. Single pass without DP overhead.

Intuit-specific tips

Intuit cares that you store prices as integer cents, not floats — if the interviewer hands you '7.25' inputs, ask whether you can convert to cents first. They also probe edge cases: an empty array (return 0), monotonically decreasing prices (return 0), and whether you can buy/sell on the same day (no — sell strictly after buy). Bonus signal: mention that this generalizes to Best Time II/III/IV with state-machine DP.

Common mistakes

  • Initializing best profit to Infinity instead of 0 — returns negative on monotonically decreasing arrays.
  • Using floats for cents — accumulates precision drift over millions of transactions.
  • Allowing buy and sell on the same day, which violates the problem constraint.

Follow-up questions

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

  • Best Time to Buy and Sell Stock II — unlimited transactions (LC 122).
  • Best Time to Buy and Sell Stock III — at most two transactions (LC 123).
  • Add a per-trade fee or cooldown day and re-derive the DP.

Solve it now

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Output

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FAQ

Why track minSoFar instead of all previous prices?

The best buy day for any sell day is always the cheapest price before it. We don't need history; just the running minimum collapses the inner loop.

How would you handle real currency inputs?

Convert to integer cents (or 4-decimal micro-units for FX) at the API boundary. The algorithm is identical; the data type prevents IEEE-754 drift across many transactions.

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