[LeetCode] 122. Best Time to Buy and Sell Stock II

🧠 Remember: “Collect every positive climb.”

The problem — 20 seconds

prices = [7, 1, 5, 3, 6, 4] → 7.

Buy and sell as many times as you like (one share at a time, sell before you rebuy). Maximize total profit.

Forget timing the market. Skim every uphill day: 1 → 5 pockets +4, 3 → 6 pockets +3. Total: 7. Downhill days cost nothing — you simply don’t trade.

7 ↘ 1 ↗ 5 ↘ 3 ↗ 6 ↘ 4
      +4         +3        → profit 7

That is the whole algorithm. Everything below just makes it obvious why.

Your first instinct

Find valleys and peaks: lowest low, highest high after it, repeat. Fiddly state-machine bookkeeping — and “only the biggest jump” fails: buying at 1 and selling at 6 skips the enforced dip at 3, the overlapping-transaction trap.

The turning point

A valley-to-peak climb is the sum of its daily up-steps: 1 → 5 as +4 equals four daily +1s. Grab each positive daily difference instead of hunting peaks — same number, three lines, no state machine.

Watch it work

Press Play. Each segment lights up day by day: green with a +profit chip for climbs, grey “skip” for slides. The counter ticks 0 → 4 → 7.

Interactive visual · Collect Every Climb

When do we collect? Every upward step.

Walk the price chart day by day. Whenever tomorrow's price is higher than today's, pocket the difference. Downhill days? Just walk past — you can't lose money you never risked. Watch the total climb to 7 on [7, 1, 5, 3, 6, 4].

total profit 0

Step 1 / 1 Speed

1profit = 0

2for i in range(len(prices) - 1): # compare day i, day i+1

3 if prices[i+1] > prices[i]: profit += prices[i+1] - prices[i]

Quick check — make it stick

When do we collect?

🧠 Memory hook: “Collect every positive climb.” Any single uphill walk equals the sum of its daily climbs, so grabbing each one is the same as timing the whole valley-to-peak perfectly.

What the code is really saying

from typing import List

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        profit = 0
        for i in range(len(prices) - 1):
            if prices[i + 1] > prices[i]:
                profit += prices[i + 1] - prices[i]
        return profit

Read it as the visual: compare today with tomorrow, pocket positive differences, ignore the rest. [1, 2, 3, 4, 5] → 4; [7, 6, 4, 3, 1] → 0 (never trade a falling market).

Why it works

Invariant: after day i, profit is the best profit using only days up to i. Any optimal valley-to-peak trade decomposes into daily up-steps summing to the same amount — exactly what the loop collects — so skimming every climb matches the optimum.

One pass, two variables: time O(n), space O(1).

🧠 Remember

Stock II → “Collect every positive climb.” Unlimited trades turn peak-picking into daily-skimming.

🔁 Try this: what changes with only one allowed trade (Stock I)? (Hint: you need the lowest price so far — the past matters now.) That’s the setup for Best Time to Buy and Sell Stock.