Sum of Matrix Diagonals

Medium Matrix

Compute the sums of the primary and secondary diagonals of a square matrix.

Sample input

[[1,2,3],[4,5,6],[7,8,9]]

Sample output

15 15

Solution

matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
n = len(matrix)
primary = sum(matrix[i][i] for i in range(n))
secondary = sum(matrix[i][n - 1 - i] for i in range(n))
print(primary, secondary)

const matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
const n = matrix.length;
let primary = 0, secondary = 0;
for (let i = 0; i < n; i++) {
  primary += matrix[i][i];
  secondary += matrix[i][n - 1 - i];
}
console.log(primary + " " + secondary);

public class Main {
    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        int n = matrix.length, primary = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            primary += matrix[i][i];
            secondary += matrix[i][n - 1 - i];
        }
        System.out.println(primary + " " + secondary);
    }
}

fun main() {
    val matrix = arrayOf(intArrayOf(1, 2, 3), intArrayOf(4, 5, 6), intArrayOf(7, 8, 9))
    val n = matrix.size
    var primary = 0
    var secondary = 0
    for (i in 0 until n) {
        primary += matrix[i][i]
        secondary += matrix[i][n - 1 - i]
    }
    println("$primary $secondary")
}

let matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
let n = matrix.count
var primary = 0, secondary = 0
for i in 0..<n {
    primary += matrix[i][i]
    secondary += matrix[i][n - 1 - i]
}
print("\(primary) \(secondary)")

void main() {
  final matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
  final n = matrix.length;
  var primary = 0, secondary = 0;
  for (var i = 0; i < n; i++) {
    primary += matrix[i][i];
    secondary += matrix[i][n - 1 - i];
  }
  print('$primary $secondary');
}

#include <iostream>
using namespace std;

int main() {
    int matrix[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int n = 3, primary = 0, secondary = 0;
    for (int i = 0; i < n; i++) {
        primary += matrix[i][i];
        secondary += matrix[i][n - 1 - i];
    }
    cout << primary << " " << secondary << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int matrix[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int n = 3, primary = 0, secondary = 0;
    for (int i = 0; i < n; i++) {
        primary += matrix[i][i];
        secondary += matrix[i][n - 1 - i];
    }
    printf("%d %d\n", primary, secondary);
    return 0;
}