Minimum Operations to Halve Array Sum - Practice Coding Problems

Minimum Operations to Halve Array Sum - Problem

You are given an array nums of positive integers representing a collection of values. Your goal is to reduce the total sum of the array by at least half using the minimum number of operations possible.

In each operation, you can:

Choose any number from the array
Replace it with exactly half of its current value
The reduced number can be chosen again in future operations

Example: If you have [5, 19, 8, 1] with sum = 33, you need to reduce the sum by at least 16.5. You could halve 19 → 9.5, then halve 9.5 → 4.75, getting sum = 18.25, which is ≤ 16.5.

Return the minimum number of operations needed to achieve this goal.

Input & Output

example_1.py — Basic Case

$ Input: [5, 19, 8, 1]

› Output: 5

💡 Note: Initial sum = 33, target = 16.5. Operations: 19→9.5 (sum=23.5), 9.5→4.75 (sum=18.75), 8→4 (sum=14.75), need to continue until sum ≤ 16.5. Total 5 operations needed.

example_2.py — Small Array

$ Input: [3, 8, 20]

› Output: 3

💡 Note: Initial sum = 31, target = 15.5. Operations: 20→10 (sum=21), 10→5 (sum=16), 8→4 (sum=12). After 3 operations, sum=12 ≤ 15.5.

example_3.py — Edge Case

$ Input: [1]

› Output: 1

💡 Note: Initial sum = 1, target = 0.5. One operation: 1→0.5. After 1 operation, sum=0.5 ≤ 0.5.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁷
Important: Elements can become fractional after operations

Visualization

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Understanding the Visualization

Setup

Build max heap with all array elements, calculate target sum = total/2

Greedy Choice

Always pop the largest element from heap (root of max heap)

Halve & Update

Halve the element, subtract reduction from sum, push halved value back

Repeat

Continue until current sum ≤ target, count operations

Key Takeaway

🎯 Key Insight: The greedy approach is optimal because halving the largest element always gives the maximum possible reduction in each operation, minimizing the total number of operations needed.

Asked in

a Amazon 45 ⊞ Microsoft 32 G Google 28 f Meta 22

The optimal solution uses a greedy approach with a max heap. Always choose the largest element to halve, as this provides maximum sum reduction per operation. The max heap efficiently maintains access to the largest element in O(log n) time, resulting in overall O(n log n) complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Simulation)	O(n^k)	O(k)	Try all possible sequences of operations until target is reached
Greedy with Max Heap (Optimal)	O(n log n)	O(n)	Always halve the largest element using a max heap for efficiency

Brute Force (Simulation) — Algorithm Steps

Calculate target sum (original sum / 2)
Use recursive backtracking to try all possible operation sequences
For each state, try halving each element in the array
Track the minimum operations needed to reach target
Return the minimum found

Visualization

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Step-by-Step Walkthrough

Initial State

Start with original array [5, 19, 8, 1], sum = 33, target = 16.5

Branch on Choices

At each step, try halving each element

Recursive Exploration

Recursively explore each branch until target is reached

Track Minimum

Keep track of minimum operations found across all branches

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int backtrack(double* arr, int n, double currentSum, int ops, double target) {
    if (currentSum <= target) return ops;
    
    int minOps = INT_MAX;
    for (int i = 0; i < n; i++) {
        double* newArr = malloc(n * sizeof(double));
        memcpy(newArr, arr, n * sizeof(double));
        double reduction = newArr[i] / 2;
        newArr[i] = reduction;
        int result = backtrack(newArr, n, currentSum - reduction, ops + 1, target);
        if (result < minOps) minOps = result;
        free(newArr);
    }
    return minOps;
}

int halveArray(int* nums, int numsSize) {
    double total = 0;
    for (int i = 0; i < numsSize; i++) total += nums[i];
    double target = total / 2;
    
    double* arr = malloc(numsSize * sizeof(double));
    for (int i = 0; i < numsSize; i++) arr[i] = nums[i];
    
    int result = backtrack(arr, numsSize, total, 0, target);
    free(arr);
    return result;
}

int main() {
    int nums[] = {5, 19, 8, 1};
    int size = 4;
    printf("%d\n", halveArray(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n^k)

Where k is the number of operations needed. Each operation has n choices, leading to exponential complexity

✓ Linear Growth

Space Complexity

O(k)

Recursion stack depth proportional to number of operations

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Simulation) — Algorithm Steps

Visualization

Code -

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