Problem Description

You have an array nums containing positive integers. You can perform operations on this array where each operation consists of:

• Selecting any number from the array
• Reducing that number to exactly half of its current value
• The halved number remains in the array and can be selected for future operations

Your goal is to find the minimum number of operations needed to reduce the total sum of the array by at least half of its original sum.

For example, if the original sum of the array is 20, you need to perform operations until the sum becomes 10 or less.

The key insight is that you're allowed to repeatedly halve the same number multiple times. Each time you halve a number, it counts as one operation, and the halved value replaces the original value in the array for potential future operations.

The solution uses a greedy approach with a max heap. Since we want to minimize the number of operations, we should always halve the largest number in the array at each step, as this gives us the maximum reduction per operation. The algorithm:

1. Calculates half of the original sum (s = sum(nums) / 2) as our target reduction
2. Uses a max heap (implemented with negative values in Python's min heap) to track array elements
3. Repeatedly extracts the maximum value, halves it, adds it back to the heap, and decrements s by the amount reduced
4. Continues until s ≤ 0, meaning we've reduced the sum by at least half
5. Returns the count of operations performed

Intuition

To minimize the number of operations, we need to maximize the reduction in sum with each operation. Since each operation halves a number, the larger the number we choose to halve, the greater the reduction we achieve.

Think about it this way: if we have numbers [10, 2, 1], halving 10 gives us a reduction of 5, while halving 2 only gives us a reduction of 1. Clearly, halving the larger number is more efficient.

This leads us to a greedy strategy: always halve the current largest number in the array. But here's the catch - after halving a number, it becomes smaller and might no longer be the largest. For instance, if we have [10, 8] and halve 10 to get 5, our array becomes [5, 8]. Now 8 is the largest, so we should halve it next.

This dynamic nature of "what's the current largest?" suggests we need a data structure that can efficiently:

1. Find the maximum element
2. Remove it
3. Insert the halved value back
4. Repeat until we've reduced the sum by at least half

A max heap (priority queue) is perfect for this because it maintains the maximum element at the top with O(log n) insertion and extraction operations.

The target is to reduce the sum by at least half, so we track how much we still need to reduce (s = original_sum / 2). Each time we halve a number x, we reduce the sum by x/2, so we subtract this from s. When s becomes zero or negative, we've achieved our goal.

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

Solution Approach

The implementation follows the greedy strategy with a max heap to always halve the largest available number:

Step 1: Calculate Target Reduction

s = sum(nums) / 2

We calculate half of the original sum. This represents how much we need to reduce from the array's total.

Step 2: Initialize Max Heap

pq = []
for x in nums:
    heappush(pq, -x)

Python's heapq module only provides a min heap, so we use a common trick: store negative values to simulate a max heap. When we push -x, the largest positive number becomes the smallest negative number, sitting at the heap's root.

Step 3: Greedy Reduction Process

ans = 0
while s > 0:
    t = -heappop(pq) / 2
    s -= t
    heappush(pq, -t)
    ans += 1

The main loop continues while we still need to reduce the sum (s > 0):

• Extract the maximum value: -heappop(pq) gives us the largest positive number
• Calculate the reduction: t = max_value / 2 represents how much we're reducing
• Update our target: s -= t decreases the remaining amount we need to reduce
• Put the halved value back: heappush(pq, -t) adds the halved number back to the heap (negated for max heap behavior)
• Count the operation: ans += 1

Time Complexity: O(n + k log n) where n is the array size and k is the number of operations. Building the initial heap takes O(n), and each of the k operations involves heap operations taking O(log n).

Space Complexity: O(n) for storing the heap with all array elements.

The algorithm guarantees finding the minimum operations because at each step, we make the locally optimal choice (halving the largest number), which leads to the globally optimal solution.

Example Walkthrough

Let's walk through a concrete example with nums = [5, 19, 8, 1].

Initial Setup:

• Original sum = 5 + 19 + 8 + 1 = 33
• Target reduction needed: s = 33 / 2 = 16.5
• Build max heap (using negatives): [-19, -8, -5, -1]

Operation 1:

• Extract max: 19 (from -19)
• Halve it: 19 / 2 = 9.5
• Reduction achieved: 9.5
• Update s: 16.5 - 9.5 = 7
• Insert 9.5 back: heap becomes [-9.5, -8, -5, -1]
• Array conceptually: [5, 9.5, 8, 1]

Operation 2:

• Extract max: 9.5
• Halve it: 9.5 / 2 = 4.75
• Reduction achieved: 4.75
• Update s: 7 - 4.75 = 2.25
• Insert 4.75 back: heap becomes [-8, -5, -4.75, -1]
• Array conceptually: [5, 4.75, 8, 1]

Operation 3:

• Extract max: 8
• Halve it: 8 / 2 = 4
• Reduction achieved: 4
• Update s: 2.25 - 4 = -1.75
• Since s ≤ 0, we've achieved our goal!

Result: 3 operations needed

The final sum would be 5 + 4.75 + 4 + 1 = 14.75, which is less than half of the original sum (16.5). By always choosing the largest number to halve, we minimized the number of operations needed.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The time complexity consists of:

• Building the initial heap by pushing all n elements: O(n × log n) since each heappush operation takes O(log n) time
• The while loop performs operations until the sum is reduced by at least half. In the worst case, we might need O(n × log n) operations total:
  • Each iteration performs one heappop operation: O(log n)
  • Each iteration performs one heappush operation: O(log n)
  • The maximum number of iterations is bounded by O(n × log n) because each operation halves a value, and in the worst case scenario (repeatedly halving the largest values), the number of operations needed is logarithmic in the ratio between the largest and smallest meaningful values

Space Complexity: O(n)

The space complexity is determined by:

• The priority queue pq stores all n elements from the input array: O(n)
• Additional variables (s, ans, t) use constant space: O(1)

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

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

Common Pitfalls

1. Integer Division Instead of Float Division

Pitfall: Using integer division (//) instead of float division (/) when calculating the target reduction or halving values. This can lead to incorrect results due to precision loss.

# WRONG - loses precision
target_reduction = sum(nums) // 2  # Integer division
halved_value = largest_element // 2  # Integer division

# CORRECT - maintains precision
target_reduction = sum(nums) / 2  # Float division
halved_value = largest_element / 2  # Float division

2. Forgetting to Negate Values for Max Heap

Pitfall: Since Python's heapq only provides a min heap, forgetting to negate values when pushing or popping can result in processing the smallest values instead of the largest ones.

# WRONG - treats it as min heap
heappush(max_heap, num)  # Pushes positive value
largest = heappop(max_heap)  # Gets smallest value

# CORRECT - simulates max heap
heappush(max_heap, -num)  # Pushes negative value
largest = -heappop(max_heap)  # Negates to get largest positive value

3. Not Putting Halved Value Back in Heap

Pitfall: After halving a number, forgetting to add it back to the heap. The problem allows selecting the same (now halved) number for future operations.

# WRONG - doesn't add halved value back
while target_reduction > 0:
    largest = -heappop(max_heap)
    target_reduction -= largest / 2
    # Missing: heappush(max_heap, -largest/2)
    operations_count += 1

# CORRECT - adds halved value back for potential future operations
while target_reduction > 0:
    largest = -heappop(max_heap)
    halved = largest / 2
    target_reduction -= halved
    heappush(max_heap, -halved)  # Put it back in heap
    operations_count += 1

4. Incorrect Target Calculation

Pitfall: Misunderstanding what "reduce by half" means. The target should be half of the original sum, not the final sum we want to achieve.

# WRONG - tries to reach half the sum as final value
current_sum = sum(nums)
target_final_sum = current_sum / 2
while current_sum > target_final_sum:
    # Complex tracking of current sum

# CORRECT - tracks how much to reduce
target_reduction = sum(nums) / 2
while target_reduction > 0:
    # Simply subtract each reduction from target

5. Using Fixed Array Instead of Heap

Pitfall: Trying to use a sorted array and re-sorting after each operation, which is inefficient.

# WRONG - O(n log n) per operation due to sorting
nums.sort(reverse=True)
while target_reduction > 0:
    nums[0] = nums[0] / 2
    target_reduction -= nums[0]
    nums.sort(reverse=True)  # Expensive re-sort
    operations_count += 1

# CORRECT - O(log n) per operation with heap
# Use heap operations as shown in the solution