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O(n) solution:

def diagonalDifference(arr):
ROWS = len(arr)
res = 0
for i in range(ROWS):
diff = (arr[i][i] - arr[i][ROWS-1-i])
res += diffreturn abs(res)

Python Brute Force Solution

For the left-to-right diagonal: Loop through all elements, sum elements where row index == column index.

For the right-to-left diagonal: Loop through all elements, sum elements where row index + column index == n-1.

def diagonalDifference(arr):
    n = len(arr)
    left_diag = 0
    right_diag = 0
    for i in range(n):
        for j in range(n):
            if i == j:
                left_diag += arr[i][j]
            if i + j == n-1:
                right_diag += arr[i][j]
    return abs(left_diag - right_diag)

Time Complexity: O(n²), two nested loops. Space Complexity: O(1)

Optimal Solution You don't need to traverse every element, each diagonal has only n elements (one per row/column).

Access the diagonals directly using index formulas in a single loop.

def diagonalDifference(arr):
    n = len(arr)
    left_diag = 0
    right_diag = 0
    for i in range(n):
        left_diag += arr[i][i]
        right_diag += arr[i][n-i-1]
    return abs(left_diag - right_diag)

Time Complexity: O(n), just one pass Space Complexity: O(1) >

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