在C++中，找到使数组所有元素相等所需的操作次数

Input : arr[] = {7, 3, 3, 3}
Output : 3

#include<bits/stdc++.h>
using namespace std;
int findOperationCount(int arr[],int n){
   int j = 0, operations = 0;
   int maxVal = arr[0];
   int minVal = arr[0];
   int maxValInd = 0;
   for (int i = 1; i < n; i++){
      if(arr[i] > maxVal){
         maxVal = arr[i];
         maxValInd = i;
      }
      if(arr[i] < minVal){
         minVal = arr[i];
      }
   }
   for (int i =0;i<n;i++){
      if (arr[i] != maxVal && arr[i] <= minVal && arr[i] != 0){
         arr[j] += 1;
         arr[maxValInd] -= 1;
         maxVal -= 1;
         operations += 1;
         j += 1;
      }
      else if (arr[i] != 0){
         j += 1;
      }
   }
   for (int i = 0; i < n; i++){
      if (arr[i] != maxVal){
         operations = -1;
         break;
      }
   }
   return operations;
}
int main(){
   int arr[] = {4, 4, 8, 4};
   int n = sizeof(arr)/sizeof(arr[0]);
   cout<<"The number of operations required to make all array  elements Equal is "<<findOperationCount(arr, n);
   return 0;
}

The number of operations required to make all array elements Equal is 3