Buy and Sell Stock IV
A hard 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. You may complete at most k transactions. Find the maximum profit achievable.
Example 1
- Input
- k = 2, prices = [2,4,1]
- Output
- 2
- Why
- Buy at 2, sell at 4 for profit 2.
Example 2
- Input
- k = 2, prices = [3,2,6,5,0,3]
- Output
- 7
- Why
- Buy at 2, sell at 6 (profit 4), buy at 0, sell at 3 (profit 3). Total = 7.
Example 3
- Input
- k = 1, prices = [1,2,3,4,5]
- Output
- 4
- Why
- Buy at 1, sell at 5 for profit 4.
Constraints
- 0 <= k <= 100
- 1 <= prices.length <= 10^5
- 0 <= prices[i] <= 1000
How to think about it
Updated 2026-09-09Generalizing to k transactions means tracking k distinct buy-and-sell stages. For each stage j from 1 to k, you either carry forward yesterday's held/sold status, buy stock j using proceeds from transaction j - 1, or sell stock j at the current price.
Approaches, worst first
Recursive trading states
time O(2^n) · space O(n * k)
Recurse with state (day, transactions_remaining, holding_boolean). Exponentially repeats subproblems across day branches without memoization, leading to redundant calculations that collapse under moderate array lengths.
2D dynamic programming grid
time O(k * n) · space O(k * n)
Maintain dp[t][d] as max profit using at most t transactions up to day d. For each t, maintain maxDiff = max(maxDiff, dp[t-1][d] - prices[d]) to compute dp[t][d] in O(1) per cell.
Rolling k-state arraysWrite this one
time O(k * n) · space O(k)
Maintain arrays buy[j] and sell[j] of length k + 1. Initialize buy to -infinity and sell to 0. On each price p, update buy[j] = max(buy[j], sell[j-1] - p) and sell[j] = max(sell[j], buy[j] + p). If 2k >= n, bypass DP and sum all positive price differences greedily.
Where people lose marks · 3
- Failing to handle k = 0, which must immediately return 0 profit.
- When k >= n / 2, you can execute as many transactions as there are price increases; failing to short-circuit to a simple O(n) greedy sweep can cause unnecessary DP overhead.
- Initializing buy array with 0 instead of -infinity, making buying appear cost-free.
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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