Removing Minimum Number of Magic Beans

/*
If we choose to make x bags empty,we should choose the x bags with the least amount
of beans.so better to sort the array.so we need to find the m which will be out beans
for non-empty bags.so possibility of m can be numbers in array because we can only
reduce the beans,can't add any beans.so after sorting,we can make empty ith bag and make
all bag after i to i+1th bean,in this way all arr[i] will be m once.



*/
long long _min(long long a,long long b){
    if(a<b) return a;
    return b;
}
class Solution {
public:
    long long minimumRemoval(vector<int>& beans) {
        sort(beans.begin(),beans.end());
        long long sum=0,prev_sum=0;
        for(auto itr:beans) sum+=itr;
        long long ans=sum;  //This is must because you can't put INT_MAX here because if k=0,then answer will be sum not INT_MAX.
        for(long long int i=0;i<beans.size();i++){
            long long temp=prev_sum+sum-prev_sum-(beans.size()-i)*beans[i];
            ans=_min(ans, temp );
            prev_sum+=beans[i];
        }
        return ans;
    }
};

#Python
class Solution:
    def minimumRemoval(self, beans: List[int]) -> int:
        arr=sorted(beans)
        total=sum(beans)
        ans=total
        n=len(beans)
        for i in range(len(arr)):
            ans=min(ans,total-((n-i)*arr[i]))
        return ans