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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 <stdio.h>
2
3int diagonalSum(int** mat, int n) {
4 int sum = 0;
5 for (int i = 0; i < n; i++) {
6 sum += mat[i][i];
7 if (i != n - i - 1) {
8 sum += mat[i][n - i - 1];
9 }
10 }
11 return sum;
12}
13
14int main() {
15 int* mat[3];
16 int row1[] = {1, 2, 3};
17 int row2[] = {4, 5, 6};
18 int row3[] = {7, 8, 9};
19 mat[0] = row1;
20 mat[1] = row2;
21 mat[2] = row3;
22 printf("%d\n", diagonalSum(mat, 3));
23 return 0;
24}The function iterates over the elements, and sums those at positions [i][i] for the primary diagonal and [i][n-i-1] for the secondary diagonal. It avoids double-counting the center element in an odd-length matrix.
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#include <vector>
int diagonalSum(const std::vector<std::vector<int>>& mat) {
int n = mat.size();
int primarySum = 0, secondarySum = 0;
for (int i = 0; i < n; i++) {
primarySum += mat[i][i];
secondarySum += mat[i][n - i - 1];
}
if (n % 2 == 1) {
primarySum -= mat[n / 2][n / 2];
}
return primarySum + secondarySum;
}
int main() {
std::vector<std::vector<int>> mat = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
std::cout << diagonalSum(mat) << std::endl;
return 0;
}This C++ solution acknowledges the potential double-counting of the center element during diagonal sum calculation and corrects it after the sum is originally calculated.