Solution Approach

The implementation uses simulation with a greedy algorithm to handle ATM operations.

Data Structure Setup:

• self.d = [20, 50, 100, 200, 500]: Array storing the denomination values in ascending order
• self.cnt = [0] * 5: Array tracking the count of each denomination (initially all zeros)
• The indices in both arrays correspond to each other - index 0 represents $20 bills, index 1 represents $50 bills, and so on

Deposit Operation: The deposit method is straightforward - we iterate through the input array banknotesCount and add each count to our internal storage:

for i, x in enumerate(banknotesCount):  self.cnt[i] += x

This simply increments our bill inventory for each denomination. Time complexity is O(1) since we always process exactly 5 denominations.

Withdraw Operation: The withdraw method implements the greedy approach with the following steps:

1. Initialize result array: ans = [0] * 5 to track how many of each bill we'll withdraw

2. Greedy allocation from largest to smallest:

   for i in reversed(range(self.m)):  ans[i] = min(amount // self.d[i], self.cnt[i])  amount -= ans[i] * self.d[i]

   • We iterate backwards through denominations (from $500 down to $20)
   • For each denomination, we calculate the maximum bills we can use: min(amount // self.d[i], self.cnt[i])
   • This takes the minimum of: bills needed to not exceed amount vs. bills available
   • We subtract the value of used bills from the remaining amount

3. Validation check:

   if amount > 0:  return [-1]

   If there's still amount left after trying all denominations, the withdrawal is impossible

4. Commit the transaction:

   for i, x in enumerate(ans):  self.cnt[i] -= x return ans

   Only if the withdrawal is valid do we actually deduct the bills from our inventory and return the withdrawal distribution

The beauty of this approach is its simplicity - by processing denominations in descending order and greedily taking as many bills as possible at each step, we naturally prioritize larger bills while ensuring we don't exceed the requested amount. The final validation step ensures we only complete transactions that can provide exact change.

Example Walkthrough

Let's walk through a concrete example to see how the ATM handles deposits and withdrawals.

Initial State:

• ATM is empty: cnt = [0, 0, 0, 0, 0] (representing counts of [$20, $50, $100, $200, $500])

Step 1: Deposit Operation

deposit([0, 0, 1, 2, 1])

• We're depositing: 0×$20, 0×$50, 1×$100, 2×$200, 1×$500
• After deposit: cnt = [0, 0, 1, 2, 1]
• Total cash in ATM: $100 + $400 + $500 = $1000

Step 2: Successful Withdrawal

withdraw(600)

Let's trace through the greedy algorithm:

1. Start with amount = 600, ans = [0, 0, 0, 0, 0]

2. Try $500 bills (index 4):

   • Can use: min(600 // 500, 1) = min(1, 1) = 1 bill
   • Take 1×$500: ans[4] = 1
   • Remaining: amount = 600 - 500 = 100

3. Try $200 bills (index 3):

   • Can use: min(100 // 200, 2) = min(0, 2) = 0 bills
   • Take 0×$200: ans[3] = 0
   • Remaining: amount = 100

4. Try $100 bills (index 2):

   • Can use: min(100 // 100, 1) = min(1, 1) = 1 bill
   • Take 1×$100: ans[2] = 1
   • Remaining: amount = 100 - 100 = 0

5. Try $50 and $20 bills: amount = 0, so we take 0 of each

6. Check: amount = 0 ✓ (exact change possible)

7. Update inventory: cnt = [0, 0, 0, 2, 0] (removed 1×$100 and 1×$500)

8. Return: [0, 0, 1, 0, 1]

Step 3: Failed Withdrawal

withdraw(600)

Now let's try to withdraw $600 again with our updated inventory:

1. Start with amount = 600, ans = [0, 0, 0, 0, 0]

2. Try $500 bills (index 4):

   • Can use: min(600 // 500, 0) = min(1, 0) = 0 bills (we have none!)
   • Remaining: amount = 600

3. Try $200 bills (index 3):

   • Can use: min(600 // 200, 2) = min(3, 2) = 2 bills
   • Take 2×$200: ans[3] = 2
   • Remaining: amount = 600 - 400 = 200

4. Try $100 bills (index 2):

   • Can use: min(200 // 100, 0) = min(2, 0) = 0 bills (we have none!)
   • Remaining: amount = 200

5. Try $50 and $20 bills: We have 0 of each, so amount stays at 200

6. Check: amount = 200 > 0 ✗ (cannot make exact change!)

7. Return: [-1] (withdrawal failed, inventory unchanged)

This example demonstrates two key aspects:

• The greedy algorithm always tries larger bills first
• If exact change cannot be made, the entire withdrawal is rejected and no bills are dispensed

Time and Space Complexity

Time Complexity:

• __init__(): O(1) - Initializes fixed-size arrays with 5 elements
• deposit(): O(1) - Iterates through a fixed array of size 5 to update counts
• withdraw(): O(1) - Performs two fixed iterations through arrays of size 5 (one reversed loop for calculation and one forward loop for updating counts)

Since all operations work with arrays of constant size (5 denominations: 20, 50, 100, 200, 500), the time complexity for all methods is O(1).

Space Complexity:

• The class maintains two fixed-size arrays: self.d with 5 denominations and self.cnt with 5 counters
• The withdraw() method creates a temporary array ans of size 5
• All space usage is constant regardless of input size

Therefore, the overall space complexity is O(1).

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

Common Pitfalls

Pitfall 1: Not Handling Transaction Atomicity

The Problem: A critical mistake is modifying the ATM's banknote counts before verifying if the withdrawal is actually possible. Consider this flawed implementation:

def withdraw(self, amount: int) -> List[int]:  withdrawal_counts = [0] * self.num_denominations   for i in reversed(range(self.num_denominations)):  notes_to_use = min(amount // self.denominations[i], self.banknote_counts[i])  withdrawal_counts[i] = notes_to_use  # WRONG: Updating counts immediately  self.banknote_counts[i] -= notes_to_use  amount -= notes_to_use * self.denominations[i]   if amount > 0:  # Problem: We've already modified the ATM state!  return [-1]   return withdrawal_counts

This breaks the atomicity principle - if the withdrawal fails, the ATM state has already been corrupted and we cannot easily roll back the changes.

The Solution: Always validate the entire transaction before modifying any state. Only update the banknote counts after confirming the withdrawal is possible:

# First, simulate the withdrawal for i in reversed(range(self.num_denominations)):  notes_to_use = min(amount // self.denominations[i], self.banknote_counts[i])  withdrawal_counts[i] = notes_to_use  amount -= notes_to_use * self.denominations[i]  # Then, check if it's valid if amount > 0:  return [-1]  # Finally, commit the changes for i, count in enumerate(withdrawal_counts):  self.banknote_counts[i] -= count

Pitfall 2: Incorrect Greedy Logic with Division

The Problem: Using floating-point division or incorrect integer division can lead to attempting to withdraw more bills than needed:

# WRONG: This might try to use more bills than necessary notes_to_use = self.banknote_counts[i] # Just taking all available bills if notes_to_use * self.denominations[i] <= amount:  withdrawal_counts[i] = notes_to_use  amount -= notes_to_use * self.denominations[i]

The Solution: Always use integer division to calculate the exact number of bills needed, then take the minimum with available bills:

# CORRECT: Calculate exact bills needed, then take minimum with available notes_to_use = min(amount // self.denominations[i], self.banknote_counts[i])

Pitfall 3: Attempting to Optimize with Backtracking

The Problem: Some might think the greedy approach could fail in cases where a different combination would work, leading to complex backtracking implementations:

# UNNECESSARY: Trying to find alternative combinations def withdraw_with_backtrack(self, amount, index=4):  if amount == 0:  return True  if index < 0 or amount < 0:  return False   # Try different combinations...  for count in range(self.banknote_counts[index], -1, -1):  if count * self.denominations[index] <= amount:  # Complex recursive logic...

Why This is Wrong: The problem specifically states that the ATM must use a greedy approach with larger denominations first. The ATM doesn't search for alternative combinations - it strictly follows the greedy algorithm. If the greedy approach fails, the withdrawal fails, even if another combination might work.

For example, if you need to withdraw $600 with available bills: 3×$200 and 1×$500:

• The ATM will use 1×$500 first (greedy), leaving$100 needed
• Even though 3×$200 would work perfectly, the ATM won't backtrack
• The withdrawal returns [-1]

The correct approach is to stick with the simple greedy algorithm as specified in the problem requirements.