Minimize count of divisions by 2 required to make all array elements equal

Minimize count of divisions by 2 required to make all array elements equal
Given an array arr[] consisting of N positive integers, the task is to find the minimum count of divisions(integer division) of array elements by 2 to make all array elements the same.
Examples:

Input: arr[] = {3, 1, 1, 3}Output: 2Explanation:Operation 1: Divide arr[0] ( = 3) by 2. The array arr[] modifies to {1, 1, 1, 3}.Operation 2: Divide arr[3] ( = 3) by 2. The array arr[] modifies to {1, 1, 1, 1}.Therefore, the count of division operations required is 2.
Input: arr[] = {2, 2, 2}Output: 0

Approach: The idea to solve the given problem is to find the maximum number to which all the elements in the array can be reduced. Follow the steps below to solve the problem:Initialize a variable, say ans, to store the minimum count of division operations required.
Initialize a HashMap, say M, to store the frequencies of array elements.
Traverse the array arr[] until any array element arr[i] is found to be greater than 0. Keep dividing arr[i] by 2 and simultaneously update the frequency of the element obtained in the Map M.
Traverse the HashMap and find the maximum element with frequency N. Store it in maxNumber.
Again, traverse the array arr[] and find the number of operations required to reduce arr[i] to maxNumber by dividing arr[i] by 2 and add the count of operations to the variable ans.
After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:

C++

#include
using namespace std;
  

void makeArrayEqual(int A[], int n)
{
    
    map mp;
  
    
    
    
    int max_number = 0;
  
    
    
    int ans = 0;
  
    
    
    for (int i = 0; i < n; i++) {            int b = A[i];                     while (b) {             mp[b]++;                             b /= 2;         }     }                  for (auto x : mp) {                              if (x.second == n) {                                          max_number = x.first;         }     }             for (int i = 0; i < n; i++) {         int b = A[i];                                       while (b != max_number) {                                          ans++;             b /= 2;         }     }             cout