Problem Statement

Best Time to Buy and Sell Stock

You get a list called prices. The number prices[i] is the cost of one stock on day i. You want to buy on one day and sell on a later day to make the most money. The catch is you have to buy first and sell after, never the other way around. Return the biggest profit you can make. If there is no way to make money, return 0.

easyArraySliding WindowSliding WindowTime: O(n) · Space: O(1)

Translate the prompt

Given a list of daily prices, return the maximum profit from buying on one day and selling on a later day. If no profitable pair exists, return 0.

Signals to notice

max over a running baselineone transactionleft-to-right sweep

Brute force first

Try every (buy, sell) pair: O(n²). Correct but recomputes minimums over and over.

The key insight

For any day i picked as the sell day, the best buy day is the cheapest day in `0..i-1`. That cheapest-so-far value can be maintained in O(1) per step as you scan.

Trace it on `prices=[7,1,5,3,6,4]`

p=7  min=7 profit=7-7=0   best=0
p=1  min=1 profit=1-1=0   best=0
p=5  min=1 profit=5-1=4   best=4
p=3  min=1 profit=3-1=2   best=4
p=6  min=1 profit=6-1=5   best=5
p=4  min=1 profit=4-1=3   best=5  → return 5

What must stay true

After processing day i, `minSoFar == min(prices[0..i])` and `best == max(prices[j] - min(prices[0..j-1]))` over j in 0..i.

Shape of the loop

minSoFar = +inf
best = 0
for p in prices:
  best = max(best, p - minSoFar)
  minSoFar = min(minSoFar, p)

Pseudocode only — the full worked solution lives in the Solution tab.

Easy way to go wrong

Updating `minSoFar` BEFORE computing today’s profit. That lets today buy and sell on the same day, producing 0 when a real answer exists on day i.

Sliding Window Pattern