Minimum Operations to Halve Array Sum
Problem
Given an array of positive integers nums, return the minimum number of operations required to reduce the sum of nums by at least half, where in each operation you can halve any element.
- 1 ≤ nums.length ≤ 10⁵
- 1 ≤ nums[i] ≤ 10⁷
Example
nums = [5,19,8,1]3The initial sum is 33, so the target is to reduce it to 16.5 or less. The algorithm uses a max heap to always halve the largest element available. First, halve 19 to 9.5 (sum reduces by 9.5), then halve 9.5 to 4.75 (sum reduces by 4.75), then halve 8 to 4 (sum reduces by 4). After these three operations, the sum is reduced by 18.25, which is more than half. The minimal number of operations is 3.
Approach
Straightforward Solution
A naive approach would repeatedly scan the array to find the largest element to halve, resulting in O(n * m) time complexity where m is the number of operations, which is inefficient for large inputs.
Core Observation
The problem requires repeatedly reducing the largest element to maximize the sum reduction per operation. This greedy approach ensures the greatest immediate decrease in sum, which is essential to minimize the total operations.
Path to Optimal
The key recognition signals are 'reduce sum by half' and 'choose any element to halve repeatedly'. These indicate a priority queue (max heap) pattern because we need efficient access to the current largest element at each step. Using a max heap allows O(log n) extraction and insertion, enabling a greedy strategy that always halves the largest element, ensuring minimal operations.
Optimal Approach
Transform the array into a max heap (using negative values for Python's min heap). Initialize the target as half the sum. In each operation, pop the largest element, halve it, push it back, and reduce the target accordingly. Continue until the target is zero or less. This guarantees the minimal number of operations with O(n + m log n) time complexity.
Time
O(n + m log n)
Building the heap takes O(n). Each operation involves a pop and push on the heap, each O(log n). The number of operations m is bounded by the number of halvings needed to reduce the sum by half.
Space
O(n)
The heap stores all n elements, requiring O(n) auxiliary space. No additional significant data structures are used.
Pattern Spotlight
Heap / Priority Queue (Greedy Maximum Extraction)
When repeatedly selecting and modifying the largest element to optimize a global metric, use a max heap to efficiently access and update the current maximum, enabling a greedy approach that minimizes total operations.
Solution
| 1 | import heapq |
| 2 | |
| 3 | class Solution: |
| 4 | def halveArray(self, nums: List[int]) -> int: |
| 5 | target = sum(nums) / 2 |
| 6 | heap = [-num for num in nums] |
| 7 | heapq.heapify(heap) |
| 8 | |
| 9 | ans = 0 |
| 10 | while target > 0: |
| 11 | ans += 1 |
| 12 | x = heapq.heappop(heap) |
| 13 | target += x / 2 |
| 14 | heapq.heappush(heap, x / 2) |
| 15 | |
| 16 | return ans |
Step-by-Step Solution
Initialize Target Sum and Max Heap from Input Array
| 5 | target = sum(nums) / 2 |
| 6 | heap = [-num for num in nums] |
| 7 | heapq.heapify(heap) |
Objective
To compute the target sum to reduce to and prepare a max heap for efficient extraction of the largest element.
Key Insight
Calculating half the total sum sets a clear stopping condition. Converting the array into a max heap (by negating values) allows O(log n) access to the largest element, which is critical for the greedy halving strategy to be efficient.
Interview Quick-Check
Core Logic
Negating values transforms Python's min heap into a max heap, enabling efficient retrieval of the largest element.
State & Boundaries
The target is set as half the sum of all elements, representing the amount of sum reduction needed.
Common Pitfalls & Bugs
Forgetting to negate values or to heapify the list results in incorrect or inefficient extraction of the largest element.
Greedily Halve Largest Elements Until Target is Met
| 9 | ans = 0 |
| 10 | while target > 0: |
| 11 | ans += 1 |
| 12 | x = heapq.heappop(heap) |
| 13 | target += x / 2 |
| 14 | heapq.heappush(heap, x / 2) |
Objective
To repeatedly halve the largest element, update the target sum, and count operations until the sum is reduced by at least half.
Key Insight
Each operation halves the largest element, which yields the maximum possible immediate reduction in sum. By pushing the halved value back into the heap, the algorithm dynamically updates the set of candidates for halving, ensuring the next operation always targets the current largest element. This greedy approach guarantees minimal operations.
Interview Quick-Check
Core Logic
At each iteration, pop the largest element, halve it, update the target by adding half the popped value (since values are negative), and push the halved value back to maintain the heap property.
State & Boundaries
The loop continues until the target is zero or less, meaning the sum has been reduced by at least half.
Common Pitfalls & Bugs
Incorrectly updating the target or pushing the wrong value back into the heap can cause infinite loops or incorrect counts.
Complexity
Each heap operation is O(log n), and the number of operations is bounded by the number of halvings needed, ensuring efficiency.
Return the Total Number of Halving Operations
| 16 | return ans |
Objective
To output the minimal count of operations needed to reduce the sum by half.
Key Insight
Returning the accumulated count after the loop terminates provides the minimal number of halving operations, as the greedy approach ensures no fewer operations can achieve the target.
Interview Quick-Check
Core Logic
The variable tracking the number of operations is incremented exactly once per halving, reflecting the minimal count.
Line Analysis
This solution has 3 Critical lines interviewers watch for.
heap = [-num for num in nums]Create a list of negated values from the input array to simulate a max heap.
Negating values allows Python's min heap to function as a max heap, enabling efficient extraction of the largest element at each step.
x = heapq.heappop(heap)Extract the current largest element (negated) from the heap.
Popping the largest element ensures the greedy choice of halving the element that yields the greatest immediate sum reduction.
target += x / 2Update the target by adding half of the popped element's value (accounting for negation).
Since the values are negated, adding half the popped value reduces the target by the correct amount, reflecting the sum decrease from halving the largest element.
target = sum(nums) / 2Calculate half of the total sum of the input array.
This value represents the target amount by which the sum must be reduced, providing a clear stopping condition for the algorithm.
heapq.heapify(heap)Transform the negated list into a heap data structure.
Heapifying the list organizes it to support O(log n) insertion and extraction operations, which is essential for the greedy strategy's efficiency.
heapq.heappush(heap, x / 2)Push the halved value (negated) back into the heap for future consideration.
Reinserting the halved element maintains the heap property and allows the algorithm to continue halving the largest available element in subsequent iterations.
while target > 0:Continue halving while the target sum reduction has not been achieved.
The loop condition ensures the algorithm runs until the sum is reduced by at least half, guaranteeing correctness.
return ansReturn the total number of halving operations performed.
Returning the accumulated count provides the minimal number of operations needed to reduce the sum by half, as guaranteed by the greedy approach.
ans += 1Increment the operation count for the current halving.
Each iteration corresponds to one halving operation, so incrementing here accurately counts total operations.
ans = 0Initialize the counter for the number of halving operations performed.
Tracking the number of operations is necessary to return the minimal count once the sum reduction target is met.
Test Your Understanding
Why does halving the largest element at each step guarantee the minimal number of operations to reduce the sum by half?
Because halving the largest element yields the greatest immediate reduction in the total sum, a greedy choice that ensures the sum decreases as quickly as possible, minimizing the total number of operations.
Related Problems
Heaps pattern
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