Buy and Sell Stock with Cooldown
A medium Dynamic Programming problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
- Platform
- LeetCode
The problem
You are given an array of daily stock prices. After selling a stock, you must wait one day before buying again (cooldown). Find the maximum profit you can achieve.
Example 1
- Input
- prices = [1,2,3,0,2]
- Output
- 3
- Why
- Buy on day 0, sell on day 1, cooldown on day 2, buy on day 3, sell on day 4. Profit = 1+2 = 3.
Example 2
- Input
- prices = [1]
- Output
- 0
- Why
- Only one day, no transaction possible.
Example 3
- Input
- prices = [1,2,4,5,7]
- Output
- 6
- Why
- Buy on day 0, sell on day 4 for profit 6.
Constraints
- 1 <= prices.length <= 5000
- 0 <= prices[i] <= 1000
How to think about it
Updated 2026-09-09On any given day you are in one of three states: holding a stock, just sold a stock (triggering cooldown), or resting. The cooldown rules dictate that you can only buy if you rested yesterday, and selling today forces tomorrow into a mandatory rest.
Approaches, worst first
Recursive state machine
time O(2^n) · space O(n)
Recurse over days carrying a holding flag and a cooldown flag, branching on act or wait at each one. Correct and close to how the rules read, but two different sequences of earlier trades can arrive at the same day in the same state, and each re-explores the whole remaining calendar from scratch.
Dynamic programming state arrays
time O(n) · space O(n)
Maintain three arrays: hold[i], sold[i], rest[i]. Transition using: hold[i] = max(hold[i-1], rest[i-1] - prices[i]); sold[i] = hold[i-1] + prices[i]; rest[i] = max(rest[i-1], sold[i-1]).
Constant space state trackersWrite this one
time O(n) · space O(1)
Replace arrays with three scalar variables: hold, sold, rest. Initialize hold = -prices[0], sold = 0, rest = 0. Iterate through prices updating the scalars in lockstep. Compressing the three history buffers into running registers retains full state fidelity while shrinking memory usage to zero auxiliary allocations.
Where people lose marks · 3
- Buying immediately after selling: transitioning to hold from sold[i-1] directly violates the mandatory 1-day cooldown.
- Seeding initial hold state with 0: hold represents buying a stock, so day 0 hold must be -prices[0]. Setting it to 0 treats buying as free.
- Returning the hold state: maximum profit must be taken from max(sold, rest), never hold.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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