Solution Approach

The solution implements a greedy algorithm using a max heap (priority queue) to always select the maximum element at each step.

Implementation Steps:

1. Initialize the Max Heap: Since Python's heapq module only provides a min heap, we simulate a max heap by negating all values. We convert the input array nums into negative values and heapify it:

   h = [-v for v in nums] heapify(h)

2. Initialize the Answer: Start with ans = 0 to accumulate our score.

3. Perform k Operations: For each of the k operations:

   • Extract the maximum element (which is the minimum negative value):

     v = -heappop(h)

   • Add this value to our total score:

     ans += v

   • Calculate the reduced value using ceil(v / 3) and push it back to the heap as a negative number:

     heappush(h, -(ceil(v / 3)))

4. Return the Result: After k operations, return the accumulated score.

Time Complexity: O(k log n) where n is the length of the array. Each heap operation (pop and push) takes O(log n) time, and we perform k such operations.

Space Complexity: O(n) for storing the heap.

The key insight is that by using a heap, we ensure that at each step we're always picking the largest available value, which guarantees the maximum possible total score after exactly k operations.

Example Walkthrough

Let's walk through a small example with nums = [5, 8, 3] and k = 4 operations.

Initial Setup:

• Create a max heap by negating values: h = [-5, -8, -3]
• After heapifying: h = [-8, -5, -3] (min heap structure with -8 at root)
• Score = 0

Operation 1:

• Extract max (pop -8): value = 8
• Add to score: score = 0 + 8 = 8
• Calculate reduced value: ceil(8/3) = 3
• Push -3 back to heap: h = [-5, -3, -3]

Operation 2:

• Extract max (pop -5): value = 5
• Add to score: score = 8 + 5 = 13
• Calculate reduced value: ceil(5/3) = 2
• Push -2 back to heap: h = [-3, -3, -2]

Operation 3:

• Extract max (pop -3): value = 3
• Add to score: score = 13 + 3 = 16
• Calculate reduced value: ceil(3/3) = 1
• Push -1 back to heap: h = [-3, -2, -1]

Operation 4:

• Extract max (pop -3): value = 3
• Add to score: score = 16 + 3 = 19
• Calculate reduced value: ceil(3/3) = 1
• Push -1 back to heap: h = [-2, -1, -1]

Final Result: Maximum score = 19

The greedy strategy ensures we always pick the largest available value. Notice how after reducing 8 to 3, it tied with the original 3 in the array, and we picked both of them in subsequent operations before picking the smaller values.

Time and Space Complexity

Time Complexity: O(n + k × log n)

• Building the initial max heap from the array takes O(n) time using heapify()
• The main loop runs k iterations
• In each iteration:
  • heappop() operation takes O(log n) time
  • heappush() operation takes O(log n) time
• Total time for the loop: k × O(log n) = O(k × log n)
• Overall time complexity: O(n) + O(k × log n) = O(n + k × log n)

Space Complexity: O(n)

• The heap h is created by storing all n elements from the input array, requiring O(n) additional space
• No other significant auxiliary space is used
• Therefore, the space complexity is O(n)

Note: The space complexity could be considered O(1) if we modify the input array in-place instead of creating a new heap, but in this implementation, a new list is created for the heap.

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

Common Pitfalls

1. Integer Division vs True Division

One of the most common mistakes is using integer division (//) instead of true division (/) when calculating the ceiling value.

Incorrect approach:

new_value = ceil(max_value // 3) # Wrong! Integer division happens first

Why it's wrong: The expression max_value // 3 performs integer division (floor division) first, then ceil() is applied to an already truncated integer, which has no effect.

Example: For max_value = 10:

• Wrong: ceil(10 // 3) = ceil(3) = 3
• Correct: ceil(10 / 3) = ceil(3.333...) = 4

Solution: Always use true division (/) before applying ceil():

new_value = ceil(max_value / 3) # Correct

2. Forgetting to Negate Values for Max Heap

Python's heapq module only provides a min heap. A common mistake is forgetting to negate values consistently when simulating a max heap.

Incorrect patterns:

# Forgetting to negate when pushing back heappush(max_heap, new_value) # Wrong! Should be -new_value  # Forgetting to negate when popping max_value = heappop(max_heap) # Wrong! Should be -heappop(max_heap)

Solution: Always maintain consistency with negation:

• Negate when pushing: heappush(max_heap, -new_value)
• Negate when popping: max_value = -heappop(max_heap)

3. Manual Ceiling Calculation Error

Some might try to implement ceiling manually instead of using math.ceil(), leading to errors:

Incorrect manual implementation:

new_value = (max_value + 2) // 3 # Works for positive integers but risky

Why it's problematic: While (n + 2) // 3 works for positive integers as a ceiling of n/3, it's:

• Less readable and maintainable
• Prone to errors if the divisor changes
• May fail for edge cases or if the problem constraints change

Solution: Use the built-in math.ceil() function for clarity and correctness:

from math import ceil new_value = ceil(max_value / 3)