Equalize the Array using map, Hackerrank O(n)

Equalize the Array

time complexity O(n) 

Easy one different approach... Learning Map

Array . If we delete  and , all of the elements in the resulting array, , will be equal. Deleting these  elements is minimal because our only other options would be to delete  elements to get an array of either  or . Thus, we print  on a new line, as that is the minimum number of deletions resulting in an array where all elements are equal.




#include<stdio.h>
#include<map>
#include <iostream>

using namespace std;

int main()
{
    int n,i;
    int arr[100];
    scanf("%d",&n);
    std::map<int,int> newmap;
    std::map<int,int>::iterator it;
    for(i=0;i<n;i++)
    {
        scanf("%d",&arr[i]);
    }
   for(i=0;i<n;i++)
    {
        if (newmap.find(arr[i]) == newmap.end())
        {

            newmap.insert ( std::pair<int,int>(arr[i],0));

        }
        else
        {
            newmap[arr[i]] ++;
        }

    }
int current_max=0;
for (it=newmap.begin(); it!=newmap.end(); ++it)
{

    if( it->second>current_max)
    {
        current_max=it->second;
    }
}

printf("%d",n-(current_max+1));
    return 0;
}