How do you solve Minimum Operations to Halve Array Sum?

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.

Minimum Operations to Halve Array Sum

Q: What is the time complexity of Minimum Operations to Halve Array Sum?

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.

Q: What pattern does Minimum Operations to Halve Array Sum use?

The Pattern: Heap / Priority Queue (Greedy Maximum Extraction) The Heuristic: 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.

Medium Heap / Priority Queue

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

Input: nums = [5,19,8,1]

Output: 3

The 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

Preview

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…

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Optimal Approach

Preview

Transform the array into a max heap (using negative values for Python's min heap). Initialize the target as half the sum…

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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

Python

1import heapq
2
3class Solution:
  def halveArray(self, nums: List[int]) -> int:
      target = sum(nums) / 2
      heap = [-num for num in nums]
      heapq.heapify(heap)
      
      ans = 0
      while target > 0:
          ans += 1
          x = heapq.heappop(heap)
          target += x / 2
          heapq.heappush(heap, x / 2)
      
      return ans

Step-by-Step Solution

Initialize Target Sum and Max Heap from Input Array

5target = sum(nums) / 2
6heap = [-num for num in nums]
7heapq.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

To repeatedly halve the largest element, update the target sum, and count operations until the sum is reduced by at least half.

Return the Total Number of Halving Operations

To output the minimal count of operations needed to reduce the sum by half.

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Line Analysis

This solution has 3 Critical lines interviewers watch for.

Line 6 Critical

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.

Line 12 Critical

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.

Line 13 Critical

target += x / 2

Update 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.

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Test Your Understanding

Why does halving the largest element at each step guarantee the minimal number of operations to reduce the sum by half?

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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`