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2 hours ago+ 0 comments

JS answer:

function miniMaxSum(arr: number[]): void {
    // Write your code here
    const min = Math.min(...arr);
    const max = Math.max(...arr);

    const sum = arr.reduce((prev, curr) => {
        return prev + curr;
    })
    
    const maxSum = sum - min;
    const minSum = sum - max;

    console.log(`${minSum} ${maxSum}`)
}

17 hours ago+ 0 comments

My pythom solutioms

def miniMaxSum(arr):
    results = []
    for i in arr:
        cp_arr = arr.copy()
        cp_arr.remove(i)
        results.append(sum(cp_arr))
    print(min(results), max(results))

23 hours ago+ 0 comments

Solution in Java (more optimised version available in the comments):

public static void miniMaxSum(List arr) { Long sumOfAll = arr.stream().mapToLong(i->i).sum();
List minMaxSumList = new ArrayList(); for(int i=0; i<=arr.size() - 1; i++) { minMaxSumList.add(sumOfAll - arr.get(i)); } Collections.sort(minMaxSumList); System.out.println(minMaxSumList.get(0)+" "+minMaxSumList.get(minMaxSumList.size()-1)); //Input: 769082435 210437958 673982045 375809214 380564127 //Output: 1640793344 2199437821 }

}

1 day ago+ 0 comments

My solution for C-Sharp(C#)

public static void Main()
{

    int[] array = Array.ConvertAll(Console.ReadLine().Split(' '), c => Convert.ToInt32(c));
    Array.Sort(array);
    long minSum = 0; long maxSum = 0;
    for (int i = 0; i < array.Length - 1; i++) {
        minSum += array[i];
    }
    for (int i = 1; i < array.Length; i++) {
        maxSum += array[i];
    }
    Console.WriteLine(minSum + " " + maxSum);

}

1 day ago+ 0 comments

function miniMaxSum(arr: number[]): void {
    // Write your code here
    const sortedArr = arr.sort((a,b) => a - b);
    const reduceValues = (data: Array<number>) => data.reduce((prev, curr) => {return           prev + curr}, 0);
    const resultMin = sortedArr.filter((num, index) => {return index < (arr.length -1)});
    const resultMax = sortedArr.filter((num, index) => {return index > 0});
    
    console.log(`${reduceValues(resultMin)} ${reduceValues(resultMax)}`)
}

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