Buy and Sell Stock with Transaction Fee
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 and a fixed transaction fee. You may complete as many transactions as you like but must pay the fee for each transaction. Find the maximum profit.
Example 1
- Input
- prices = [1,3,2,8,4,9], fee = 2
- Output
- 8
- Why
- Buy at 1, sell at 8 (profit 5), buy at 4, sell at 9 (profit 3). Total = 8.
Example 2
- Input
- prices = [1,3,7,5,10,3], fee = 3
- Output
- 6
- Why
- Buy at 1, sell at 7 (profit 3), buy at 5, sell at 10 (profit 2). Total = 6.
Example 3
- Input
- prices = [1,3,2,8,4,9], fee = 10
- Output
- 0
- Why
- The fee is too high to make any profitable transaction.
Constraints
- 1 <= prices.length <= 5 * 10^4
- 1 <= prices[i] <= 5 * 10^4
- 0 <= fee <= 5 * 10^4
How to think about it
Updated 2026-09-09Deduct the fee either when buying or when selling (pick one consistently). At every day, you either hold stock or hold cash. Holding stock means you kept yesterday's stock or bought today from cash; holding cash means you kept yesterday's cash or sold today's stock minus the fee.
Approaches, worst first
Recursive trading decisions
time O(2^n) · space O(n)
Branch on buying, holding, or selling at every price point. Explores duplicate subproblems at exponential cost, lacking memoization across repeated day and balance states to prune identical branches.
2D state DP table
time O(n) · space O(n)
Table dp[i][0] for not holding and dp[i][1] for holding stock. Transitions: dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i] - fee), dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i]).
Two-variable state trackingWrite this one
time O(n) · space O(1)
Use two scalar variables `cash` = 0 and `hold` = -prices[0]. At each price: cash = max(cash, hold + price - fee); hold = max(hold, cash - price).
Where people lose marks · 3
- Paying the transaction fee twice: deducting the fee during both the buy and sell steps.
- Overwriting `cash` before using its previous value to compute `hold`. Using updated cash in the same step allows buying and selling on the exact same day while paying the fee.
- Failing to account for fee greater than price fluctuations: when fee exceeds profits, the algorithm must prefer holding cash (returning 0).
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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