public static int diagonalDifference(List> arr) { // Write your code here

int primaryDiagnolSum =0;
int secondaryDiagnolSum =0;

for(int i=1; i<=arr.size();i++) { List intArray = arr.get(i-1); primaryDiagnolSum+=intArray.get(i-1); secondaryDiagnolSum+= intArray.get(arr.size()-i);

}

int difference = primaryDiagnolSum-secondaryDiagnolSum; return (difference<0)?difference*-1: difference;

}

Java 8 solution

public static int diagonalDifference(List<List<Integer>> arr) {
// Write your code here
        int sumElementLeftToRightDiagonal= IntStream
            .range(0,arr.size())
            .mapToObj(i->arr.get(i).get(i))
            .collect(toList())
            .stream()
            .reduce(0,Integer::sum);

    int sumElementRightToLeftDiagonal=IntStream
            .range(0, arr.size())
            .mapToObj(i->arr.get(i).get(arr.size()-1-i))
            .collect(toList())
            .stream()
            .reduce(0,Integer::sum);

    return Math.abs(sumElementLeftToRightDiagonal-sumElementRightToLeftDiagonal);

}

Python solution

from functools import reduce
import operator

def diagonalDifference(arr):
    # Write your code here
    size = len(arr)
    return abs(reduce(operator.add, [arr[x][x] - arr[x][size-1-x] for x in range(size)]))

My solution in C++

int diagonalDifference(vector<vector<int>> arr) {
int left_result = 0;
int right_result = 0;
int diagonalDiff = 0; 
for (int i = 0; i < arr.size(); i++){
     
  left_result += arr[i][i];
  right_result += arr[i][arr.size()-i-1];
}

diagonalDiff = left_result - right_result;
return abs(diagonalDiff);
}

My submission is Below: Please give inputs about the approach

 public static int diagonalDifference(List<List<Integer>> arr) {
    // Write your code here
    int countp=0,counts =0;
        for(int i=0,j=arr.size()-1;i<arr.size()&&j>=0;i++,j--){
            countp = countp + arr.get(i).get(i);
            counts= counts + arr.get(i).get(j);
        }
        if(countp<=counts){
            return counts-countp;
        }else{
            return countp-counts;
        }
    }