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This approach involves traversing the matrix and accessing diagonal elements using indices.
Primary Diagonal: For an element to be on the primary diagonal, its row index must be equal to its column index (i.e., mat[i][i]).
Secondary Diagonal: For an element to be on the secondary diagonal, the sum of its row index and column index must be equal to n-1 (i.e., mat[i][n-i-1]).
Time Complexity: O(n) since we traverse the matrix diagonals once.
Space Complexity: O(1) as we use a constant amount of extra space.
1def diagonalSum(mat):
2 n = len(mat)
3 total_sum = 0
4 for i in range(n):
5 total_sum += mat[i][i]
6 if i != n - i - 1:
7 total_sum += mat[i][n - i - 1]
8 return total_sum
9
10# Example usage
11matrix = [
12 [1, 2, 3],
13 [4, 5, 6],
14 [7, 8, 9]
15]
16print(diagonalSum(matrix))This Python function calculates the diagonal sum by iterating through each index of the square matrix and adding the appropriate diagonal elements, taking care to not double-count the middle element in odd-length matrices.
An alternative method is to calculate both diagonal sums and subtract the repeated center element if it exists. This approaches the same goal in a slightly different way by not thinking too much about the double-count case upfront during the main loop.
Time Complexity: O(n) because of the loop through the matrix diagonals.
Space Complexity: O(1) as we use constant additional space.
1
In this Java solution, the matrix diagonals are individually summed, and the repeated center element is subtracted if the matrix has an odd width and height, ensuring no element is counted twice.