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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.
1#include <iostream>
2#include <vector>
3
4int diagonalSum(const std::vector<std::vector<int>>& mat) {
5 int n = mat.size();
6 int sum = 0;
7 for (int i = 0; i < n; i++) {
8 sum += mat[i][i];
9 if (i != n - i - 1) {
10 sum += mat[i][n - i - 1];
11 }
12 }
13 return sum;
14}
15
16int main() {
17 std::vector<std::vector<int>> mat = {
18 {1, 2, 3},
19 {4, 5, 6},
20 {7, 8, 9}
21 };
22 std::cout << diagonalSum(mat) << std::endl;
23 return 0;
24}The solution calculates the sum of diagonals by iterating over the matrix elements. It adds the element at position [i][i] for the primary diagonal and [i][n-i-1] for the secondary diagonal, ignoring the middle element if n is odd.
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
This approach calculates the primary and secondary diagonal sums separately. After the sums are obtained, it checks if the matrix is odd in size; if so, it subtracts the middle element from the total sum to correct the double counting.