# Diagonal Difference Hacker Rank Solution Best & Easiest Diagonal Difference Hacker Rank Solution Best & Easiest

Given a square matrix, calculate the absolute difference between the sums of its diagonals.

or example, the square matrix arr is shown below:

``````1 2 3
4 5 6
9 8 9  ``````

The left-to-right diagonal = 1 + 5 +9 = 15. The right to left diagonal = 3 + 5 + 9 = 17. Their absolute difference is |15 – 17| = 2.

Function description

Complete the DiagonalDifference function in the editor below.

diagonalDifference takes the following parameter:

• int arr[n][m]: an array of integers

Return

• int: the absolute diagonal difference

## Input Format

The first line contains a single integer, n, the number of rows and columns in the square matrix arr.
Each of the next n lines describes a row, arr[i], and consists of n space-separated integers arr[i][j].

## Constraints

• -100 <= arr[i][j] <= 100

## Output Format

Return the absolute difference between the sums of the matrix’s two diagonals as a single integer.

Sample Input

``````3
11 2 4
4 5 6
10 8 -12``````

Sample Output

``15``

Explanation

The primary diagonal is:

``````11
5
-12``````

Sum across the primary diagonal: 11 + 5 – 12 = 4

The secondary diagonal is:

``````     4
5
10``````

Sum across the secondary diagonal: 4 + 5 + 10 = 19
Difference: |4 – 19| = 15

Note: |x| is the absolute value of x

## Diagonal Difference Hacker Rank Solution:

### Problem solution in C++ programming

```int diagonalDifference(vector<vector<int>> arr) {

size_t size = arr.size();
int sum{};
int i=0;

for(const auto& vec : arr)
sum+= (vec[size-1-i] - vec[i++]);

return abs(sum);
}
```

### Problem solution in JavaScript programming

```function diagonalDifference(arr) {
let first_diagonal = 0, second_diagonal = 0
let len = arr.length - 1
for(let i=0; i<arr.length; i++){
first_diagonal += arr[i][i];
second_diagonal += arr[i][len-i];
}
let res = first_diagonal - second_diagonal
return Math.abs(res)
}
```

### Problem solution in Python programming

```def diagonalDifference(arr):      Write your code here
first_diagonal = second_diagonal = 0
# Time: O(n)
# Space: O(1)
for i in range(len(arr)):
first_diagonal += arr[i][i]
second_diagonal += arr[i][len(arr) - 1 - i]
return abs(first_diagonal - second_diagonal)
```