This might be simple in python

def min_max_sum(arr): # Write your code here1 2 3 4 result = [] for i in range(len(arr)): org = arr[:] org.pop(i) result.append(sum(org)) return f"{min(result)} {max(result)}"

if name == 'main':

arr = list(map(int, input().rstrip().split()))
b = min_max_sum(arr)
print(b)

public static void miniMaxSum(List<Integer> arr) {
// Write your code here

Collections.sort(arr);
//System.out.println(arr.toString());
Long max = 0L;
Long min = 0L;
for(int i = 0; i < arr.size() ; i ++){
     if(i != 0){
        max = max + arr.get(i);
     }
     if(i != (arr.size() -1)){
        min = min + arr.get(i);
     }
}
System.out.println(min +" "+ max);
}

python code, simple and easy to understand

sumTotal = sum(arr)
minI = min(arr)
maxI = max(arr)
maxNum = sumTotal-minI
minNum = sumTotal-maxI
print(minNum, maxNum)

Java Solution Easy to understand public static void miniMaxSum(List arr) {

Collections.sort(arr);

long minSum = 0;
long maxSum =0;


for(int i = 0; i < arr.size() -1; i++ ){
    minSum += arr.get(i);
}

for(int i = 1; i < arr.size() ; i++ ){
    maxSum += arr.get(i);
}

System.out.println(minSum + " " + maxSum);
}

// java code

public static void miniMaxSum(List<Integer> arr) {
        long max = 0;
        long min = 0;
        
        for(int i = 0; i < arr.size(); i++) {
            long count = 0;
            
            for (int j = 0; j < arr.size(); j++) {
                if (j != i) {
                    count += arr.get(j);
                }
            }
            
            if (count > max || max == 0) max = count;
            if (count < min || min == 0) min = count;
        }
        
        System.out.printf("%d %d", min, max);
    }