My c++ solution:

int diagonalDifference(vector> arr) { int diag1 = 0; int diag2 = 0; for(int i = 0; i

for(int i = 0; i<arr.size(); i++){
    for(int j = 0; j<arr.size(); j++){
        if(i+j==arr.size()-1){
            diag2 += arr[i][j];
        }else{
            continue;
        }
    }
}
int diff = diag2 - diag1;
if(diag2 - diag1 <0){
    return -(diff);
}else{
    return diff;
}

}

My C# solution:

public static int diagonalDifference(List<List<int>> arr)
{
        var diagonalSumA = 0;
        var diagonalSumB = 0;
        var rowCounter = 0;

        foreach (var item in arr) {
            for (int i = 0; i < item.Count; i++) {
                if (rowCounter == i) {
                    diagonalSumA += item[i];
                }

                if (i == (-rowCounter + (item.Count - 1))) {
                    diagonalSumB += item[i];
                }
            }

            rowCounter++;
        }

        return Math.Abs(diagonalSumA - diagonalSumB);
}[](https://)

For Javascript solution:

function diagonalDifference(arr) {
    let tampung = 0;
    let tampung2 = 0;
    let n = arr.length;
    for(let i=0; i < n; i++){
        tampung = tampung + arr[i][i];
        tampung2 = tampung2 + arr[i][n - 1 - i];
    }
    return Math.abs(tampung - tampung2);
}

My Java Solution:

    int rowSize = arr.size();
    int leftSum = IntStream.range(0, rowSize)
                           .map(i -> arr.get(i).get(i))
                           .sum();
    int rightSum = IntStream.range(0, rowSize)
                            .map(i -> arr.get(i).get(rowSize - (1 + i)))
                            .sum();
    return Math.abs(leftSum - rightSum);

**My Shortest Python Solution **

def diagonalDifference(arr):
    L = R = 0
    n = len(arr[0])
    for i in range(n):
        L += arr[i][i]
        R += arr[i][n-1-i ]
    return abs(L-R)
		
		or else :
		

def diagonalDifference(n,arr):
L = R = 0
for i in range(n):
L += arr[i][i]
R += arr[i][n-1-i ]
return abs(L-R)

result = diagonalDifference(n,arr)