# Diagonal difference

Practicing in HackerRank I found this interesting exercise.

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

For 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 2.

## My solution

```
#!/bin/python3
import math
import os
import random
import re
import sys
#
# Complete the 'diagonal_difference' function below.
#
# The function is expected to return an INTEGER.
# The function accepts 2D_INTEGER_ARRAY arr as parameter.
#
def diagonal_difference(arr:list)->int:
'''
Calc the absolute difference.
Calc the sum of diagonal elements from left to right (sum1),
Calc the sum of diagonal elements from right to left (sum2),
Calc the absolute difference sum1-sum2.
Parameters
----------
arr : list of int
Square List of integers.
Return
------
abs_diff: int.
Examples
--------
arr = [[1, 3, 6], [4, 1, 7], [7, 3, 6]]
abs_diff = 6
arr = [[1, 3, 6, 3], [4, 1, 7, 3], [7, 3, 6, 4], [8, 2, 5, 3]]
abs_diff = 10
'''
sum1 = 0
sum2 = 0
for i in range(N):
sum1 += arr[i][i]
sum2 += arr[i][N-i-1]
abs_diff=abs(sum1-sum2)
return abs_diff
if __name__ == '__main__':
N = 4
source = [[1, 3, 6, 3], [4, 1, 7, 3], [7, 3, 6, 4], [8, 2, 5, 3]]
x = diagonal_difference(source)
print(x)
```

The key to this exercise is how to loop through the list, in my case, I decided calc both sums in the same loop, but each sum can be calculated in different loops.

```
for i in range(N):
sum1 += arr[i][i]
sum2 += arr[i][N-i-1]
```

Other important point, if we don't have N, it's easy, simply use the method "len":

```
for i in range(len(arr)):
# code
```

Thanks for reading :)

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