通过将任何元素减半来最小化将数组总和减半的操作

给定一个数组Arr[]，任务是找出使数组元素之和小于或等于其初始值一半的最小操作数。在一种这样的操作中，允许将任何数组元素的值减半。

例子：

Input: Arr[] = [4, 6, 3, 9, 10, 2]
Output: 5
Explanation: The initial sum = (4+6+3+9+10+2) = 34
1st step: choose 10 and make it 5, sum = 29
2nd step: choose 9 and make it 4, sum = 24
3rd step: choose 6 and make it 3, sum = 21
4th step: choose 5 and make it 2, sum =18
5th step: choose 4 and make it 2, sum =16

Input: Arr[] = [1, 5, 8, 19]
Output: 3

编程需要懂一点英语

方法：解决问题的思路是使用堆数据结构。

The target is to reduce the number of operations to make the sum of the array half, thus,

At every operation the amount of reduction of the sum value should be as high as possible.
To achieve that, try to choose the maximum value available on the array and reducing it to its half value.
The best way to find the maximum value at every iteration is to use a priority_queue as a max-heap as max-heap will store the maximum value of the array at the top of the max-heap.

编程需要懂一点英语

要实施此方法，请按照以下步骤操作：

通过遍历数组来计算数组元素的总和。
使用priority_queue初始化一个最大堆以存储数组的所有元素。
初始化一个计数器变量为0，这个变量将存储最少的操作次数。
由于最大堆的顶部将始终保存数组中存在的最大元素，因此删除顶部元素，使其减半（整数除法）并将新值输入最大堆。
继续上一步，直到元素的总和小于或等于其初始值。

下面是上述方法的实现：

C++

// C++ code for the above approach:
  
#include 
using namespace std;
  
// Function to find minimum operations
int minops(vector& nums)
{
    int sum = 0;
    for (auto x : nums)
        sum += x;
  
    // Initializing max heap
    priority_queue pq;
    for (auto x : nums) {
        pq.push(x);
    }
    double temp = sum;
    int cnt = 0;
    while (temp > sum / 2) {
        int x = pq.top();
        pq.pop();
        temp -= ceil(x * 1.0 / 2);
        pq.push(x / 2);
        cnt++;
    }
  
    // Return count
    return cnt;
}
  
// Driver code
int main()
{
  
    vector nums = { 4, 6, 3, 9, 10, 2 };
    int count = minops(nums);
    cout << count << endl;
}

输出

时间复杂度： O(N*logN)
辅助空间： O(N)