log(n) in Python:

def miniMaxSum(arr):
    # Write your code here
    

   max_element = max(arr)
   min_element = min(arr)
   total = sum(arr)
   print(total-max_element,total-min_element)

Shortest solution in Python: def miniMaxSum(arr): # Write your code here

arr.sort()
print(sum(arr[:-1]),sum(arr[1:]))

Simple Javascript solution here, works by sorting the initial array and then reducing the first 4 indices, then last 4 indices to return the min and max sum

function miniMaxSum(arr) { // Write your code here let sorted = [...arr.sort((a, b)=>{return a-b})]; console.log(sorted.slice(0, 4).reduce((a, b)=>{ return a+b }, 0), sorted.slice(1).reduce((a, b)=>{ return a+b }, 0)) }

Java:

long summ=0,suml=0; Collections.sort(arr); for(int i=0;i<4;i++){ summ+=arr.get(i); } for(int i=arr.size()-1;i>arr.size()-5;i--){ suml+=arr.get(i); } System.out.println(summ+" "+suml); }

c++ short solution using only functions from "algorithm" library

#define all(a) (a).begin(),(a).end()
#define ll long long

void miniMaxSum(vector<int> arr) {
    ll sum = accumulate(all(arr),0ll);
    int mn = *min_element(all(arr));
    int mx = *max_element(all(arr));
    cout << sum-mx << " " << sum-mn << '\n';
}```