LeetCode 2208 - Minimum Operations to Halve Array Sum
This problem asks us to repeatedly reduce numbers in an array until the total sum of the array has been reduced by at least half, while using the minimum number of operations.
Difficulty: 🟡 Medium
Topics: Array, Greedy, Heap (Priority Queue)
Solution
Problem Understanding
This problem asks us to repeatedly reduce numbers in an array until the total sum of the array has been reduced by at least half, while using the minimum number of operations.
In one operation, we may choose any number currently in the array and replace it with exactly half of its value. Importantly, the reduced number remains in the array and can be selected again later. This means that a number such as 20 can become 10, then 5, then 2.5, and so on.
The input is an integer array nums, where every value is positive. We are asked to return the smallest number of halving operations needed so that the reduction in the total sum is at least half of the original sum.
Another way to think about the problem is this:
- Let
original_sumbe the sum of all numbers. - We want to decrease the sum by at least
original_sum / 2. - Every operation contributes some reduction amount.
- We want to minimize the number of operations.
For example, if the original sum is 40, then we need to remove at least 20 from the total. If halving a number x reduces the sum by x / 2, then choosing larger numbers gives a larger immediate reduction.
Understanding the Constraints
The constraints are large:
1 <= nums.length <= 10^51 <= nums[i] <= 10^7
A naive simulation that repeatedly scans the entire array for the largest number would be too slow because each operation might require an O(n) search, and many operations could be needed.
The constraints strongly suggest that we need an efficient way to repeatedly retrieve and update the largest element. This is a classic signal for using a heap (priority queue).
Important Edge Cases
One important edge case is when the array contains only a single element, such as [10]. Since halving 10 once reduces the total by 5, exactly half the original sum, the answer is 1.
Another case is when all numbers are equal, such as [4,4,4,4]. A naive implementation might repeatedly choose arbitrary elements inefficiently, while the optimal solution should consistently pick the largest available value.
A particularly tricky case is when one number dominates the array, such as [10000000,1,1,1]. Repeatedly halving the largest number is clearly optimal, and the implementation must efficiently handle repeated updates to the same value.
The problem guarantees that all numbers are positive integers. This guarantee matters because every halving operation always reduces the sum by a positive amount, meaning progress toward the target is guaranteed.
Approaches
Brute Force Approach
A brute-force solution would repeatedly do the following:
- Find the largest number in the array by scanning the entire array.
- Halve it.
- Update the running reduction.
- Continue until at least half the original sum has been removed.
This approach is correct because choosing the largest available value always maximizes the reduction gained from a single operation. However, repeatedly scanning the array to find the maximum is expensive.
Suppose there are k operations. Finding the largest element costs O(n) each time, leading to a total complexity of O(k * n). Since n can be 10^5, this quickly becomes too slow.
Key Insight
The key observation is that the best move at every step is to halve the largest available number.
Why?
Halving a number x decreases the total sum by x / 2. Therefore, larger numbers produce larger reductions in a single operation.
This gives us a greedy strategy:
Always halve the current maximum value.
The remaining challenge is efficiently retrieving and updating the maximum number after every operation.
A max heap solves this problem:
- Extracting the current largest value takes
O(log n). - Re-inserting its halved value also takes
O(log n).
Python provides a min heap, so we simulate a max heap using negative numbers. In Go, we implement a custom max heap using container/heap.
Approach Comparison
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(k·n) | O(1) | Repeatedly scans array to find largest value |
| Optimal | O((n + k) log n) | O(n) | Uses max heap to efficiently track largest element |
Here, k represents the number of operations performed.
Algorithm Walkthrough
Optimal Greedy + Heap Algorithm
- Compute the total sum of the array and determine the target reduction.
Let total_sum be the sum of all elements. We need to reduce the total by at least total_sum / 2.
2. Build a max heap containing all numbers.
Since we repeatedly need the largest number, a heap allows us to retrieve it efficiently. In Python, we push negative values into a min heap to simulate max heap behavior. 3. Maintain a running amount of reduction.
Instead of recalculating the array sum after every operation, we track how much total reduction has been accumulated. 4. Repeatedly remove the largest value from the heap.
This greedy choice maximizes the reduction gained in the current operation. 5. Halve the chosen number.
If the largest value is x, then:
- The reduction gained is
x / 2 - The updated value becomes
x / 2
- Add the reduction to the running total.
We continue until the accumulated reduction reaches at least total_sum / 2.
7. Push the halved number back into the heap.
Since the number can be selected again later, it must remain available. 8. Count each operation.
Once enough reduction has been achieved, return the operation count.
Why it works
The correctness comes from a greedy property.
At any step, halving a number x reduces the total sum by x / 2. Since larger x always produces a larger reduction, choosing anything other than the largest available value cannot be better in terms of minimizing operations.
The heap guarantees that we always select the best available choice efficiently. Because each operation maximizes immediate progress toward the target, the algorithm achieves the minimum number of operations.
Python Solution
from typing import List
import heapq
class Solution:
def halveArray(self, nums: List[int]) -> int:
total_sum = sum(nums)
target_reduction = total_sum / 2
max_heap = [-num for num in nums]
heapq.heapify(max_heap)
reduction = 0.0
operations = 0
while reduction < target_reduction:
largest = -heapq.heappop(max_heap)
halved = largest / 2
reduction += halved
heapq.heappush(max_heap, -halved)
operations += 1
return operations
The implementation starts by computing the total sum and the required reduction target. Since we only care about reducing the sum by at least half, we do not need to repeatedly recompute the full array sum.
A max heap is created using negative numbers because Python's heapq module only supports min heaps. By storing -num, the smallest negative value corresponds to the largest original number.
Inside the loop, the largest element is extracted and halved. Since halving a value x reduces the sum by x / 2, we add that amount directly to reduction.
The halved value is inserted back into the heap because it can be selected again in future operations. The loop continues until the required reduction threshold has been met.
Go Solution
package main
import "container/heap"
type MaxHeap []float64
func (h MaxHeap) Len() int {
return len(h)
}
func (h MaxHeap) Less(i, j int) bool {
return h[i] > h[j]
}
func (h MaxHeap) Swap(i, j int) {
h[i], h[j] = h[j], h[i]
}
func (h *MaxHeap) Push(x interface{}) {
*h = append(*h, x.(float64))
}
func (h *MaxHeap) Pop() interface{} {
old := *h
n := len(old)
item := old[n-1]
*h = old[:n-1]
return item
}
func halveArray(nums []int) int {
totalSum := 0.0
h := &MaxHeap{}
for _, num := range nums {
totalSum += float64(num)
*h = append(*h, float64(num))
}
heap.Init(h)
targetReduction := totalSum / 2
reduction := 0.0
operations := 0
for reduction < targetReduction {
largest := heap.Pop(h).(float64)
halved := largest / 2
reduction += halved
heap.Push(h, halved)
operations++
}
return operations
}
The Go implementation follows the same logic but requires explicitly implementing the heap interface from container/heap.
Unlike Python, Go does not provide a built-in max heap, so we define a custom MaxHeap type and reverse the comparison in Less() by using > instead of <.
Since values become fractional after halving, float64 is necessary. Using integers would lose precision and produce incorrect results.
Worked Examples
Example 1
Input:
nums = [5,19,8,1]
Initial sum:
33
Target reduction:
33 / 2 = 16.5
Initial max heap:
[19, 8, 5, 1]
| Operation | Largest Picked | Halved Value | Reduction Gained | Total Reduction | Heap After |
|---|---|---|---|---|---|
| 1 | 19 | 9.5 | 9.5 | 9.5 | [9.5, 8, 5, 1] |
| 2 | 9.5 | 4.75 | 4.75 | 14.25 | [8, 5, 4.75, 1] |
| 3 | 8 | 4 | 4 | 18.25 | [5, 4.75, 4, 1] |
At this point:
18.25 >= 16.5
So the answer is:
3
Example 2
Input:
nums = [3,8,20]
Initial sum:
31
Target reduction:
15.5
Initial heap:
[20, 8, 3]
| Operation | Largest Picked | Halved Value | Reduction Gained | Total Reduction | Heap After |
|---|---|---|---|---|---|
| 1 | 20 | 10 | 10 | 10 | [10, 8, 3] |
| 2 | 10 | 5 | 5 | 15 | [8, 5, 3] |
| 3 | 8 | 4 | 4 | 19 | [5, 4, 3] |
Since:
19 >= 15.5
The minimum number of operations is:
3
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O((n + k) log n) | Heap construction takes O(n), each operation performs one pop and one push |
| Space | O(n) | Heap stores all array elements |
Here, k is the number of halving operations performed.
The heap is initialized in linear time using heapify. Every operation removes the largest element and reinserts its halved version, each costing O(log n).
Although k is not explicitly bounded in the problem statement, each operation significantly reduces a number, meaning the process converges quickly in practice.
Test Cases
solution = Solution()
assert solution.halveArray([5, 19, 8, 1]) == 3 # Example 1
assert solution.halveArray([3, 8, 20]) == 3 # Example 2
assert solution.halveArray([10]) == 1 # Single element
assert solution.halveArray([1]) == 1 # Smallest valid input
assert solution.halveArray([4, 4, 4, 4]) == 4 # Equal elements
assert solution.halveArray([10000000, 1, 1, 1]) == 1 # Dominant large value
assert solution.halveArray([1, 1, 1, 1]) == 4 # Need all elements reduced
assert solution.halveArray([8, 8]) == 2 # Multiple equal max values
assert solution.halveArray([10000000]) == 1 # Maximum value boundary
assert solution.halveArray([2, 2, 2]) == 3 # Repeated balanced reductions
Test Summary
| Test | Why |
|---|---|
[5,19,8,1] |
Validates first example |
[3,8,20] |
Validates second example |
[10] |
Tests single element case |
[1] |
Tests minimum input values |
[4,4,4,4] |
Tests equal numbers |
[10000000,1,1,1] |
Tests dominant large element |
[1,1,1,1] |
Tests repeated reductions |
[8,8] |
Tests tied maximum values |
[10000000] |
Tests upper value boundary |
[2,2,2] |
Tests repeated balanced choices |
Edge Cases
Single Element Array
An input such as [10] is important because there is only one possible choice. A buggy implementation might overcomplicate the logic or fail to terminate correctly. Since halving 10 immediately reduces the sum by 5, which is exactly half the original total, the algorithm correctly returns 1.
Extremely Large Dominating Value
An array like [10000000,1,1,1] tests whether the algorithm properly prioritizes the largest number. A poor strategy might waste operations reducing small numbers first. The heap-based greedy approach always chooses the maximum, immediately halving 10000000, which already exceeds the required reduction threshold.
Repeated Selection of the Same Number
A case such as [20,1,1] is subtle because the same number should be selected multiple times. After halving 20 into 10, it may still remain the largest value and should be selected again. Some incorrect implementations forget to push the halved value back into the heap, producing wrong results. This solution reinserts every halved number, ensuring repeated selections work correctly.
Equal Values
Arrays such as [4,4,4,4] ensure the implementation handles ties correctly. Since every number provides the same reduction, the order does not matter. The heap naturally handles equal priorities, and the algorithm still produces the minimum number of operations.