How do you solve Minimum Operations to Halve Array Sum on LeetCode?

Minimum Operations to Halve Array Sum (LeetCode #2208) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Always halve the largest number to maximize reduction in sum.

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

The optimal solution for Minimum Operations to Halve Array Sum runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) because we build the max-heap in O(n) time and each operation of extracting and inserting into the heap takes O(log n).

What is the brute force solution for Minimum Operations to Halve Array Sum?

The brute force approach for Minimum Operations to Halve Array Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we repeatedly halve the largest number in the array until the sum is reduced by at least half. This method is straightforward but inefficient as it may require many operations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Halve Array Sum?

Minimum Operations to Halve Array Sum uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Heap, Greedy.

What are the interview tips for Minimum Operations to Halve Array Sum?

Explain your thought process clearly, especially why you choose a certain data structure. Practice explaining the time and space complexities of your solution. Consider edge cases, such as arrays with very small or very large numbers.

What are common mistakes when solving Minimum Operations to Halve Array Sum?

Failing to use a max-heap, leading to inefficient solutions. Not correctly updating the current sum after halving.

#2208

Minimum Operations to Halve Array Sum

Medium

ArrayGreedyHeap (Priority Queue)HeapGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a max-heap (priority queue) to efficiently retrieve and halve the largest number in each operation. This approach minimizes the number of operations needed to reduce the sum by at least half.

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Algorithm

6 steps

1Step 1: Calculate the initial sum of the array.
2Step 2: Create a max-heap from the array elements.
3Step 3: While the current sum is greater than half of the initial sum, extract the maximum element from the heap.
4Step 4: Halve this maximum element and update the current sum.
5Step 5: Insert the halved value back into the heap and increment the operation count.
6Step 6: Return the count of operations.

solution.py16 lines

1# Full working Python code
2import heapq
3
4def min_operations(nums):
5    initial_sum = sum(nums)
6    target_sum = initial_sum / 2
7    current_sum = initial_sum
8    operations = 0
9    max_heap = [-num for num in nums]
10    heapq.heapify(max_heap)
11    while current_sum > target_sum:
12        max_num = -heapq.heappop(max_heap)
13        current_sum -= max_num / 2
14        heapq.heappush(max_heap, -max_num / 2)
15        operations += 1
16    return operations

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Complexity note: The time complexity is O(n log n) because we build the max-heap in O(n) time and each operation of extracting and inserting into the heap takes O(log n).

1Always halve the largest number to maximize reduction in sum.
2Using a max-heap allows for efficient retrieval of the largest element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.