#2208

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Operations to Halve Array Sum

Minimize operations to halve an array's sum using greedy choices and heap data structures.

Problem Statement

You are given an array of positive integers. In one operation, you can choose any number from the array and reduce it to exactly half. The same number can be selected in future operations. Your task is to return the minimum number of operations needed to reduce the sum of the array by at least half.

For example, if the array is [5, 19, 8, 1], the sum is initially 33. You can reduce the sum by choosing elements and halving them. The objective is to achieve a sum of 16.5 or less with the fewest halving operations.

Examples

Example 1

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

Output: 3

The initial sum of nums is equal to 5 + 19 + 8 + 1 = 33. The following is one of the ways to reduce the sum by at least half: Pick the number 19 and reduce it to 9.5. Pick the number 9.5 and reduce it to 4.75. Pick the number 8 and reduce it to 4. The final array is [5, 4.75, 4, 1] with a total sum of 5 + 4.75 + 4 + 1 = 14.75. The sum of nums has been reduced by 33 - 14.75 = 18.25, which is at least half of the initial sum, 18.25 >= 33/2 = 16.5. Overall, 3 operations were used so we return 3. It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Example 2

Input: nums = [3,8,20]

Output: 3

The initial sum of nums is equal to 3 + 8 + 20 = 31. The following is one of the ways to reduce the sum by at least half: Pick the number 20 and reduce it to 10. Pick the number 10 and reduce it to 5. Pick the number 3 and reduce it to 1.5. The final array is [1.5, 8, 5] with a total sum of 1.5 + 8 + 5 = 14.5. The sum of nums has been reduced by 31 - 14.5 = 16.5, which is at least half of the initial sum, 16.5 >= 31/2 = 15.5. Overall, 3 operations were used so we return 3. It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 107

Solution Approach

Greedy Approach Using a Max Heap

To minimize the number of operations, always halve the largest number in the array first. A max heap is used to efficiently retrieve and update the largest element after each operation. This ensures that the largest number gets halved first, reducing the total sum as efficiently as possible.

Step-by-Step Halving

In each operation, the largest number is halved, and the new value is reinserted into the heap. Continue this process until the total sum of the array is reduced by at least half. Each operation effectively reduces the sum by halving one element at a time.

Efficient Heap Management

The max heap allows for efficient retrieval and updating of the largest number in logarithmic time, making it well-suited for this problem where frequent element removals and insertions are required. This ensures the solution is optimized for both time and space complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n log n), where n is the number of elements in the array. This is due to the need to insert and remove elements from the heap, each of which takes logarithmic time. The space complexity is O(n) for storing the heap.

What Interviewers Usually Probe

The candidate should understand greedy algorithms and their application to real-world problems like this one.
The candidate should be able to justify using a heap for efficiently finding and updating the largest element.
The candidate should highlight the importance of reducing the sum quickly by halving the largest element at each step.

Common Pitfalls or Variants

Common pitfalls

Not using a heap may result in inefficient solutions, especially with large arrays.
Forgetting to account for floating-point precision when halving elements multiple times.
Using a non-optimal strategy for halving elements, such as randomly choosing numbers, can lead to more operations.

Follow-up variants

What if the array contains zeros or negative numbers?
How would the solution change if we were allowed to halve numbers more than once per operation?
Can this approach be applied to arrays where the sum is not an integer?

FAQ

What is the optimal approach for solving the Minimum Operations to Halve Array Sum?

The optimal approach is using a greedy strategy with a max heap to always halve the largest number in the array first, minimizing operations.

Why is a max heap used in the solution?

A max heap allows for efficient retrieval and updating of the largest element, which is crucial for minimizing the number of halving operations.

How do you ensure that the sum is reduced by at least half?

By halving the largest number first, we ensure that each operation reduces the sum as quickly as possible until it is at least halved.

What are some common mistakes when solving this problem?

A common mistake is not using a heap, which can lead to inefficient solutions. Also, ignoring floating-point precision issues during halving operations can cause incorrect results.

How does GhostInterview help with solving this problem?

GhostInterview provides a step-by-step breakdown of the solution, focusing on efficient strategies like greedy algorithms and heap usage, reinforcing key concepts for interview success.

terminal

Solution

Solution 1: Greedy + Priority Queue (Max Heap)

According to the problem description, each operation will halve a number in the array. To minimize the number of operations that reduce the array sum by at least half, each operation should halve the current maximum value in the array.

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class Solution:
    def halveArray(self, nums: List[int]) -> int:
        s = sum(nums) / 2
        pq = []
        for x in nums:
            heappush(pq, -x)
        ans = 0
        while s > 0:
            t = -heappop(pq) / 2
            s -= t
            heappush(pq, -t)
            ans += 1
        return ans

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