Problem Description

You are given an array prices where prices[i] represents the stock price on day i, and an integer k representing the maximum number of transactions you can make.

You can perform two types of transactions:

1. Normal transaction: Buy stock on day i and sell it on a later day j (where i < j). Your profit is prices[j] - prices[i].

2. Short selling transaction: Sell stock on day i (without owning it first) and buy it back on a later day j (where i < j). Your profit is prices[i] - prices[j].

The key constraints are:

• You can make at most k transactions total
• You must complete one transaction before starting another (no overlapping transactions)
• You cannot buy or sell on the same day that you're completing a previous transaction (selling or buying back)

The goal is to find the maximum total profit you can achieve with at most k transactions.

For example, if prices = [3, 2, 6, 5, 0, 3] and k = 2:

• You could buy at price 2, sell at price 6 (profit = 4)
• Then short sell at price 5, buy back at price 0 (profit = 5)
• Total profit = 9

The problem uses dynamic programming with three states at each point:

• State 0: Not holding any position (neither long nor short)
• State 1: Holding a long position (bought stock)
• State 2: Holding a short position (sold stock without owning)

The solution tracks f[i][j][k] representing the maximum profit on day i with at most j transactions in state k, and uses state transitions to compute the optimal profit.

Intuition

The key insight is recognizing that at any given moment, we can be in one of three distinct states: holding nothing, holding a long position (bought stock), or holding a short position (sold stock we don't own).

Think about it this way - when we're trying to maximize profit, we need to track what position we currently hold because our next action depends on it. If we're holding a stock, we can only sell it. If we're holding a short position, we can only buy back. If we're holding nothing, we can either buy (start a long position) or sell (start a short position).

This naturally leads to a dynamic programming approach where we track three states:

• State 0: Neutral (no position)
• State 1: Long (holding stock)
• State 2: Short (owing stock)

For each day and each possible number of transactions used, we want to know: "What's the maximum profit I can have if I'm in each of these states?"

The transitions between states follow logical rules:

• To be neutral today, I could have been neutral yesterday, or I closed a position today (sold my long or bought back my short)
• To be long today, I either was already long yesterday, or I was neutral yesterday and bought today (this starts a new transaction)
• To be short today, I either was already short yesterday, or I was neutral yesterday and sold today (this also starts a new transaction)

The brilliant part is that starting a new transaction (going from neutral to either long or short) consumes one of our k allowed transactions. This is why when we transition from state 0 to state 1 or 2, we use f[i-1][j-1][0] - we're using up one transaction count.

By building up this table day by day and transaction by transaction, we eventually find the maximum profit possible with at most k transactions, which will be when we end in the neutral state (we don't want to be holding any position at the end).

Learn more about Dynamic Programming patterns.

Solution Approach

We implement a 3D dynamic programming solution where f[i][j][k] represents the maximum profit on day i with at most j transactions in state k.

State Definition:

• k = 0: Not holding any position (neutral)
• k = 1: Holding a long position (bought stock)
• k = 2: Holding a short position (sold stock without owning)

Initialization: We create a 3D array f with dimensions [n][k+1][3] where n is the number of days. For day 0, we initialize:

• f[0][j][1] = -prices[0] for all j from 1 to k (buying stock on day 0)
• f[0][j][2] = prices[0] for all j from 1 to k (short selling on day 0)
• f[0][j][0] = 0 (implicitly, as the array is initialized with zeros)

State Transitions: For each day i from 1 to n-1 and each transaction count j from 1 to k, we update the states:

1. Neutral State (k=0):

   f[i][j][0] = max(
       f[i-1][j][0],           # Stay neutral
       f[i-1][j][1] + prices[i], # Sell long position
       f[i-1][j][2] - prices[i]  # Buy back short position
   )

   We take the maximum profit from three scenarios: remaining neutral, closing a long position (selling at today's price), or closing a short position (buying back at today's price).

2. Long State (k=1):

   f[i][j][1] = max(
       f[i-1][j][1],              # Stay long
       f[i-1][j-1][0] - prices[i] # Buy stock (new transaction)
   )

   Either we were already long, or we start a new transaction by buying stock. Note that buying uses j-1 transactions since we're starting a new one.

3. Short State (k=2):

   f[i][j][2] = max(
       f[i-1][j][2],              # Stay short
       f[i-1][j-1][0] + prices[i] # Sell stock (new transaction)
   )

   Either we were already short, or we start a new transaction by short selling. This also uses j-1 transactions.

Final Result: After processing all days and transactions, we return f[n-1][k][0], which represents the maximum profit after at most k transactions, ending in a neutral position (not holding any stock or short position).

The time complexity is O(n * k) where n is the number of days and k is the maximum number of transactions. The space complexity is O(n * k) for the 3D DP array.

Example Walkthrough

Let's walk through a small example with prices = [3, 2, 6, 5] and k = 2 (maximum 2 transactions).

We'll build our DP table f[i][j][state] where:

• i = day (0 to 3)
• j = number of transactions used (0 to 2)
• state = 0 (neutral), 1 (long), or 2 (short)

Day 0 (price = 3):

• With 0 transactions: All states = 0 (can't do anything)
• With 1 transaction available:
  • f[0][1][0] = 0 (neutral, no action taken)
  • f[0][1][1] = -3 (bought at 3, profit = -3)
  • f[0][1][2] = 3 (short sold at 3, profit = 3)
• With 2 transactions available: Same as 1 transaction (can only use one on day 0)

Day 1 (price = 2):

• With 1 transaction:
  • f[1][1][0] = max(0, -3+2, 3-2) = 1 (best: close short from day 0)
  • f[1][1][1] = max(-3, 0-2) = -2 (buy at better price today)
  • f[1][1][2] = max(3, 0+2) = 3 (keep short from day 0)
• With 2 transactions:
  • f[1][2][0] = max(0, -3+2, 3-2) = 1
  • f[1][2][1] = max(-3, 1-2) = -1 (can use profit from first transaction)
  • f[1][2][2] = max(3, 1+2) = 3

Day 2 (price = 6):

• With 1 transaction:
  • f[2][1][0] = max(1, -2+6, 3-6) = 4 (sell long position for profit)
  • f[2][1][1] = max(-2, 0-6) = -2
  • f[2][1][2] = max(3, 0+6) = 6 (short sell at high price)
• With 2 transactions:
  • f[2][2][0] = max(1, -1+6, 3-6) = 5 (sell long from better entry)
  • f[2][2][1] = max(-1, 4-6) = -1
  • f[2][2][2] = max(3, 4+6) = 10 (short after completing first transaction)

Day 3 (price = 5):

• With 2 transactions:
  • f[3][2][0] = max(5, -1+5, 10-5) = 5
  • Final answer: f[3][2][0] = 5

The maximum profit is 5, achieved by:

1. Short selling at 3, buying back at 2 (profit = 1)
2. Buying at 2, selling at 6 (profit = 4) Total: 1 + 4 = 5

Solution Implementation

Time and Space Complexity

The time complexity is O(n × k), where n is the length of the array prices and k is the maximum number of transactions. This is because we have two nested loops: the outer loop iterates through all n prices, and the inner loop iterates through all k transactions. For each combination of (i, j), we perform constant time operations to update the three states.

The space complexity is O(n × k), as we create a 3D array f with dimensions [n][k+1][3]. The first dimension has size n for each day, the second dimension has size k+1 for the number of transactions (from 0 to k), and the third dimension has a fixed size of 3 for the three possible states. Since the third dimension is constant, the overall space complexity simplifies to O(n × k).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Transaction Counting

A critical pitfall is misunderstanding when a transaction is counted. Many developers incorrectly assume that both buying and selling count as separate transactions, or that both opening and closing a position count as two transactions.

The Issue: In this problem, a transaction is counted when you open a new position (either buying for long or selling for short), not when you close it. This means:

• Buying stock (opening long) uses one transaction
• Selling that stock (closing long) doesn't use an additional transaction
• Short selling (opening short) uses one transaction
• Buying back (closing short) doesn't use an additional transaction

Incorrect Implementation:

# WRONG: Counting transaction when closing positions
dp[day][transaction_count][0] = max(
    dp[day - 1][transaction_count][0],
    dp[day - 1][transaction_count - 1][1] + prices[day],  # Wrong! Selling shouldn't use a transaction
    dp[day - 1][transaction_count - 1][2] - prices[day]   # Wrong! Buying back shouldn't use a transaction
)

Correct Implementation:

# CORRECT: Only count transactions when opening positions
dp[day][transaction_count][0] = max(
    dp[day - 1][transaction_count][0],
    dp[day - 1][transaction_count][1] + prices[day],      # Closing long - no transaction used
    dp[day - 1][transaction_count][2] - prices[day]       # Closing short - no transaction used
)

dp[day][transaction_count][1] = max(
    dp[day - 1][transaction_count][1],
    dp[day - 1][transaction_count - 1][0] - prices[day]   # Opening long - uses a transaction
)

dp[day][transaction_count][2] = max(
    dp[day - 1][transaction_count][2],
    dp[day - 1][transaction_count - 1][0] + prices[day]   # Opening short - uses a transaction
)

2. Array Index Out of Bounds

When accessing dp[day - 1][transaction_count - 1][0] for buying or short selling, if transaction_count is 0, this will cause an index error or incorrect behavior.

The Issue: The loop should start from transaction_count = 1, not 0, since you need at least 1 transaction to make any trades.

Solution: Always ensure the transaction loop starts from 1:

for transaction_count in range(1, k + 1):  # Start from 1, not 0
    # ... state transitions

3. Forgetting Edge Cases

Not handling edge cases like k = 0 (no transactions allowed) or n = 0 (empty price array).

Solution: Add edge case checks at the beginning:

def maximumProfit(self, prices: List[int], k: int) -> int:
    if not prices or k == 0:
        return 0
    n = len(prices)
    # ... rest of the code