Java

Approach : 1) Add all the elements of list 2) for finding minimum sum , subtract maximum element of list from sum 3) for finding maximum sum, subtract minimum element of list from sum

double sum=0;

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

double min = sum - Collections.max(arr);
double max = sum - Collections.min(arr);

System.out.printf("%.0f ", min);
System.out.printf("%.0f", max);
}

    long total = arr.Select( a => (long)a).Sum();
    long minimum = total - arr.Max();
    long maximum = total - arr.Min();
    Console.WriteLine($"{minimum} {maximum}");

Python solution: maxSum =sum(arr) - arr[0] minSum =sum(arr) - arr[0]

for index in range(0,len(arr)):
    currentSum = sum(arr) - arr[index]
    if(currentSum > maxSum):
        maxSum = currentSum
    if(currentSum<minSum):
        minSum = currentSum

print(minSum,maxSum)

In Golang:

func miniMaxSum(arr []int32) {
    sort.SliceStable(arr, func(i, j int) bool { return arr[i]<arr[j]})
    var sum int64 = 0 
    for i := range arr{
        sum += int64(arr[i])
    }
    fmt.Println(sum-int64(arr[len(arr)-1]), sum-int64(arr[0]))
}

Java: Collections.sort(arr); long min=0; long max=0; for(int i=0;i<4;i++){ min+=arr.get(i); } for(int j=4;j>=1;j--){ max+=arr.get(j); } System.out.println(min+" "+ max);