What is the brute force approach and complexity for Best Time to Buy and Sell Stock with Cooldown?

On each day, try buy, keep, or sell; a sale jumps two days ahead with time complexity O(2^N) and space complexity O(N).

What is the optimal approach and complexity for Best Time to Buy and Sell Stock with Cooldown?

Track holding, just sold, and free to buy; update all three each day with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Best Time to Buy and Sell Stock with Cooldown?

Look for key signals in the prompt: "cannot buy on the next day" indicates A sale creates a cooldown day. The day after a sale is a different state from a normal free day.; "as many transactions as you like" indicates Many trades, so the state of each day is what we keep, not the list of trades.; "hold at most one share" indicates Holding is one state with one purchase price in it..

Best Time to Buy and Sell Stock with Cooldown

Medium·LeetCode #309 Pro OnlySlight Variation

LeetCode

Problem Statement

You are given prices, where prices[i] is the price of one share on day i. You may buy and sell as many times as you like, but you hold at most one share at a time.

There is one rule: after you sell, you cannot buy on the next day. That next day is a cooldown day.

Return the largest profit you can make.

Input

An integer array prices.

Output

The largest total profit, following the cooldown rule.

Example 1

Input: prices = [1, 2, 3, 0, 2]

Output: 3

Buy at 1 (day 0), sell at 2 (day 1). Day 2 is the cooldown. Buy at 0 (day 3), sell at 2 (day 4). The profit is (2 − 1) + (2 − 0) = 3.

Example 2

Input: prices = [1]

Output: 0

There is only one day, so no trade is possible.

Constraints

• 1 ≤ prices.length ≤ 5000
• 0 ≤ prices[i] ≤ 1000

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“cannot buy on the next day”

→ A sale creates a cooldown day. The day after a sale is a different state from a normal free day.

“as many transactions as you like”

→ Many trades, so the state of each day is what we keep, not the list of trades.

“hold at most one share”

→ Holding is one state with one purchase price in it.

Approach Complexity Ladder

Brute Force — recursion with a jump after a saleO(2^N)O(N)

On each day, try buy, keep, or sell; a sale jumps two days ahead

Better — memoized recursion (day, holding)O(N)O(N)

Cache each (day, holding) pair so it is solved only once

Optimal — three running valuesO(N)O(1)

Track holding, just sold, and free to buy; update all three each day

Brute Force Intuition

Recursion on the day and whether we hold a share. Buying or skipping moves to the next day. Selling moves to two days later, because the next day is the cooldown. Keep the best path.

Recognition rule for this sub-pattern:

Walk the days carrying the best value for each state (holding / not holding), or the best choice so far from each side.

Remember this · DP — Kadane / Subarray

Trigger

Maximum (or minimum) sum/product of a contiguous subarray, stock buy/sell, circular arrays.

Think

Is the best subarray ending at i either "i alone" or "best ending at i−1, plus i"?

Key question

Should I extend or restart?

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