Problem Description

You have several piles of stones, represented by an array piles where piles[i] is the number of stones in the i-th pile. You also have a number k which represents how many operations you must perform.

In each operation, you can:

• Pick any pile of stones
• Remove exactly floor(piles[i] / 2) stones from that pile
• The pile is updated with the remaining stones

Key points to understand:

• You must perform exactly k operations (no more, no less)
• You can choose the same pile multiple times across different operations
• floor(x) means rounding down to the nearest integer (e.g., floor(7/2) = floor(3.5) = 3)
• After removing stones, the pile has piles[i] - floor(piles[i] / 2) stones remaining, which equals ceil(piles[i] / 2)

Your goal is to minimize the total number of stones remaining across all piles after performing all k operations.

For example, if you have a pile with 9 stones and perform one operation on it:

• You remove floor(9/2) = 4 stones
• The pile now has 9 - 4 = 5 stones remaining

The strategy involves using a greedy approach with a max heap. Since you want to minimize the total remaining stones, you should always remove stones from the largest pile available, as this removes the maximum number of stones per operation. The solution uses a priority queue to efficiently track and update the largest pile after each operation.

Intuition

To minimize the total number of stones remaining, we need to maximize the number of stones we remove. Since each operation removes floor(pile[i] / 2) stones from a chosen pile, we want to remove as many stones as possible with each operation.

Consider this: removing half the stones from a larger pile removes more stones than removing half from a smaller pile. For example:

• Removing from a pile of 10 stones: we remove floor(10/2) = 5 stones
• Removing from a pile of 4 stones: we remove floor(4/2) = 2 stones

This suggests a greedy strategy: always choose the largest pile for each operation.

But why is this greedy approach optimal? Think about it this way - if we have two operations to perform and two piles with sizes a and b where a > b, we could either:

1. Remove from a twice
2. Remove from b twice
3. Remove from each once

Since the removal amount is proportional to the pile size (we remove half), and larger numbers lose more when halved, repeatedly targeting the largest pile ensures we're always removing the maximum possible stones at each step.

The challenge is that after each operation, the "largest pile" might change. If we remove stones from a pile of 10 (leaving 5), and there's another pile with 7 stones, then 7 becomes our new largest pile for the next operation.

This dynamic nature of finding the maximum after each update naturally leads us to use a max heap (priority queue). The heap allows us to:

1. Quickly find the largest pile in O(log n) time
2. Update it and maintain the heap property efficiently
3. Repeat this process k times

By always selecting and halving the current maximum pile, we ensure that we're removing the maximum possible stones across all k operations, thus minimizing the total remaining stones.

Learn more about Greedy and Heap (Priority Queue) patterns.

Solution Approach

The solution implements the greedy strategy using a max heap (priority queue) to efficiently track and update the largest pile after each operation.

Implementation Steps:

1. Create a Max Heap: Python's heapq module provides a min heap by default. To simulate a max heap, we negate all values when inserting them into the heap:

   pq = [-x for x in piles]
   heapify(pq)

   This converts the array into a heap structure where the smallest (most negative) value is at the root, which corresponds to the largest positive value.

2. Perform k Operations: For each of the k operations, we:

   • Extract the top element (largest pile in terms of absolute value)
   • Calculate the new pile size after removing half the stones
   • Put the updated pile back into the heap

   The code uses heapreplace() which is an optimized combination of heappop() and heappush():

   for _ in range(k):
       heapreplace(pq, pq[0] // 2)

   Here's what happens in each iteration:

   • pq[0] is the root (smallest negative number, representing the largest pile)
   • pq[0] // 2 performs integer division on the negative number
   • For example, if pq[0] = -9, then pq[0] // 2 = -5 (since -9 // 2 = -5 in Python)
   • This correctly represents halving the pile and keeping the ceiling value

3. Calculate the Result: After all operations, sum all elements in the heap and negate to get the positive total:

   return -sum(pq)

Why This Works:

The integer division behavior in Python is crucial here. When dividing negative numbers:

• -9 // 2 = -5 (not -4)
• This means we keep 5 stones in the pile (the ceiling of 9/2)
• We effectively removed floor(9/2) = 4 stones

Time Complexity: O(k log n) where n is the number of piles

• Each heap operation takes O(log n) time
• We perform k such operations

Space Complexity: O(n) for storing the heap

Example Walkthrough

Let's walk through a concrete example with piles = [9, 7, 4] and k = 3 operations.

Initial Setup:

• We create a max heap by negating values: pq = [-9, -7, -4]
• After heapify: pq = [-9, -7, -4] (already a valid heap)
• Total stones initially: 9 + 7 + 4 = 20

Operation 1:

• Current heap: [-9, -7, -4]
• Extract largest pile: -9 (represents pile of 9 stones)
• Remove floor(9/2) = 4 stones, leaving 5 stones
• Update heap with -5: heapreplace(pq, -9 // 2) → -9 // 2 = -5
• New heap after rebalancing: [-7, -5, -4]
• Stones removed: 4

Operation 2:

• Current heap: [-7, -5, -4]
• Extract largest pile: -7 (represents pile of 7 stones)
• Remove floor(7/2) = 3 stones, leaving 4 stones
• Update heap with -4: heapreplace(pq, -7 // 2) → -7 // 2 = -4
• New heap after rebalancing: [-5, -4, -4]
• Stones removed: 3

Operation 3:

• Current heap: [-5, -4, -4]
• Extract largest pile: -5 (represents pile of 5 stones)
• Remove floor(5/2) = 2 stones, leaving 3 stones
• Update heap with -3: heapreplace(pq, -5 // 2) → -5 // 2 = -3
• Final heap: [-4, -4, -3]
• Stones removed: 2

Final Result:

• Final heap: [-4, -4, -3]
• Sum of negated values: -4 + -4 + -3 = -11
• Return -sum(pq) = -(-11) = 11
• Total stones remaining: 4 + 4 + 3 = 11

We successfully removed 9 stones total (4 + 3 + 2) from the original 20, leaving 11 stones remaining. Notice how the greedy approach always selected the largest available pile, maximizing stone removal at each step.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + k × log n), where n is the length of the array piles. This breaks down as follows:

• O(n) for creating the negated list from piles
• O(n) for the heapify operation to build the initial max heap
• O(k × log n) for performing k iterations of heapreplace, where each heapreplace operation takes O(log n) time to maintain the heap property
• O(n) for computing the final sum

The dominant term is O(n + k × log n).

The space complexity is O(n), which is required to store the heap pq that contains all n elements from the original piles array.

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

Common Pitfalls

1. Misunderstanding Integer Division with Negative Numbers

The most critical pitfall is not understanding how Python handles integer division with negative numbers. Many developers might incorrectly think they need to manually calculate the ceiling value.

Incorrect Approach:

# Wrong: Trying to manually calculate remaining stones
for _ in range(k):
    largest = -heappop(max_heap)  # Get the largest pile
    removed = largest // 2         # Calculate stones to remove
    remaining = largest - removed  # Calculate remaining stones
    heappush(max_heap, -remaining) # Push back to heap

Why it's wrong: This adds unnecessary complexity and is less efficient than using heapreplace().

Correct Approach:

# Correct: Let Python's integer division handle it
for _ in range(k):
    heapreplace(max_heap, max_heap[0] // 2)

2. Using Min Heap Without Negation

Another common mistake is trying to use Python's min heap directly without negating values, leading to incorrect results.

Incorrect Approach:

# Wrong: Using min heap without negation
min_heap = piles[:]
heapify(min_heap)
for _ in range(k):
    smallest = heappop(min_heap)  # This gets the SMALLEST pile!
    heappush(min_heap, (smallest + 1) // 2)

Why it's wrong: This removes stones from the smallest piles first, which is the opposite of what we want.

Correct Approach:

# Correct: Negate values to simulate max heap
max_heap = [-pile for pile in piles]
heapify(max_heap)

3. Forgetting to Negate the Final Sum

After performing operations on the negated heap, forgetting to negate the sum back leads to incorrect (negative) results.

Incorrect Approach:

# Wrong: Forgetting to negate the final sum
return sum(max_heap)  # Returns a negative number!

Correct Approach:

# Correct: Negate the sum to get positive result
return -sum(max_heap)

4. Inefficient Heap Operations

Using separate heappop() and heappush() operations instead of the more efficient heapreplace().

Less Efficient Approach:

# Works but less efficient
for _ in range(k):
    largest = heappop(max_heap)
    heappush(max_heap, largest // 2)

More Efficient Approach:

# Better: heapreplace combines pop and push in one operation
for _ in range(k):
    heapreplace(max_heap, max_heap[0] // 2)

heapreplace() is more efficient because it performs the pop and push in a single operation, maintaining the heap property with fewer comparisons.