5. Best Time to Buy and Sell Stock
easyAsked at SnapFind the maximum profit from a single buy/sell on a price-per-day array. Snap uses this to verify you can convert a naive nested loop into a single-pass min-tracking sweep.
By Sam K., Founder, InterviewChamp.AI · Last verified
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Public interview reports confirming this problem appears in Snap loops.
- Glassdoor (2026-Q1)— Common Snap new-grad warm-up.
- Blind (2025-09)— Frequently paired with Two Sum to test pattern-recognition speed.
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. If no profit is possible, return 0.
Constraints
1 <= prices.length <= 10^50 <= prices[i] <= 10^4
Examples
Example 1
prices = [7,1,5,3,6,4]5Explanation: Buy at 1, sell at 6.
Example 2
prices = [7,6,4,3,1]0Explanation: Prices only decrease; no profit.
Approaches
1. Brute-force every pair
For every i, scan j > i and track max(prices[j] - prices[i]).
- 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 — too slow for 10^5 inputs. Show this only as the motivating baseline.
2. Single-pass min-tracking (optimal)
Walk left to right keeping the lowest price seen so far. At each step, candidate profit is current - min. Update both as you go.
- 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: Linear time, constant space. The 'min so far' is the recurring DP-on-prefix pattern that recurs across many Snap problems.
Snap-specific tips
Snap rewards candidates who articulate the prefix-min pattern out loud. Bonus signal: connect it to streaming analytics — Snap engineers think of this kind of single-pass scan as the same shape they use to compute running statistics on Stories view-counts.
Common mistakes
- Initializing best to -Infinity instead of 0 — the problem guarantees non-negative profit, return 0 when no trade is profitable.
- Updating minSoFar AFTER computing the candidate profit — that allows selling and buying on the same day.
- Trying to track maxSoFar instead of minSoFar — works but conceptually awkward.
Follow-up questions
An interviewer at Snap may pivot to one of these next:
- Allow unlimited transactions (LC 122).
- Allow at most two transactions (LC 123).
- Allow k transactions (LC 188) — DP table required.
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FAQ
Why use else if instead of two separate checks?
If we just updated minSoFar to today's price, then today - minSoFar == 0, which can't beat the running best. The else-if simply skips a redundant comparison. Both versions are correct.
Does the array need to be modified?
No — we only read prices. The algorithm is read-only with O(1) extra space, which is part of what makes it the textbook optimal.