Here’s a function to calculate the maximum profit considering the transaction fee:

def max_profit_with_fee(k, prices, fee):
  """
  This function calculates the maximum profit achievable with k transactions
  and a transaction fee.

  Args:
      k: Maximum allowed transactions.
      prices: Array of stock prices.
      fee: Transaction fee as a percentage.

  Returns:
      Maximum achievable profit.
  """
  # Handle edge cases
  if not prices or k == 0:
    return 0
  
  n = len(prices)
  # Initialize arrays to store maximum profit at each day
  # under two conditions: holding stock or not holding
  # (cash) considering k transactions
  dp = [[0 for _ in range(2)] for _ in range(k + 1)]

  # No transaction on day 0 (both cash and holding states)
  dp[0][0] = dp[0][1] = 0

  # Iterate through prices for each transaction (0 to k)
  for i in range(1, n):
    for j in range(k, 0, -1):
      # Cash on day i: 
      #  - Either stay in cash (previous cash)
      #  - Or sell stock from previous holding state (considering fee)
      dp[j][0] = max(dp[j][0], dp[j][1] + prices[i] - fee * prices[i] / 100)

      # Holding stock on day i:
      #  - Either stay holding (previous holding)
      #  - Or buy stock from previous cash state
      dp[j][1] = max(dp[j][1], dp[j - 1][0] - prices[i])

  # Return the maximum profit achievable with k transactions (cash state)
  return dp[k][0]

Explanation:

1. The function takes k, prices, and fee as input.
2. It handles edge cases where there are no prices or no transactions allowed (k=0).
3. It creates a 2D array dp of size (k+1) x 2. This array will store the maximum profit achievable with k transactions at each day (i) under two conditions: holding a stock (dp[j][1]) or having cash (dp[j][0]).
4. It initializes dp[0][0] and dp[0][1] to 0 as there are no transactions on day 0.
5. It iterates through the prices for each transaction (from k down to 1).
6. Inside the loop, it iterates through each day (i).
7. For the cash state (dp[j][0]), it considers two options:
   • Remain in cash (previous day’s cash value).
   • Sell the stock held previously (considering the transaction fee).
8. For the holding state (dp[j][1]), it considers two options:
   • Remain holding the stock (previous day’s holding value).
   • Buy the stock from the previous day’s cash.
9. It updates dp[j][0] and dp[j][1] with the maximum of the two options for each state.
10. Finally, it returns dp[k][0], which represents the maximum achievable profit with k transactions (cash state) after considering all days and transaction options.

This approach uses dynamic programming to efficiently calculate the maximum profit considering the transaction fee and the constraint on the number of transactions.