JavaScript program to efficiently calculate the sum of matrix diagonals

We will write a program in JavaScript to efficiently calculate the sum of the diagonals of a matrix. To do this, we will use a loop structure to iterate over the matrix and add elements located at positions corresponding to the diagonals. By exploiting the mathematical properties of matrices, we can minimize the amount of calculations required to sum the diagonals. With this approach we will be able to handle matrices of various sizes in a computationally efficient manner.

method

• To calculate the sum of the diagonals of a matrix, we need to add the element values ​​on the main diagonal (top left to bottom right) and the secondary diagonal (top right to bottom - left)

• You can use a double loop approach where one loop goes through the rows and the second loop goes through the columns to access the elements on the diagonal.

• We can keep two variables to store the sum of elements on the main diagonal and sub-diagonal respectively.

• To access elements on the main diagonal, we need to add the current row index and column index, while for elements on the sub-diagonal, we need to subtract the column index from the row index.

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• Finally, we return the sum of the two variables as the result, which will give the sum of the elements on both diagonals of the matrix.

Example

This is an example of a JavaScript program that efficiently calculates the sum of the diagonals of a matrix -

function diagonalSum(matrix) {
   let sum = 0;
   let n = matrix.length;
    
   for (let i = 0; i < n; i++) {
      sum += matrix[i][i];
      sum += matrix[i][n - i - 1];
   }
     
   if (n % 2 !== 0) {
      let mid = Math.floor(n / 2);
      sum -= matrix[mid][mid];
   }
     
   return sum;
}
const matrix = [[1, 2, 3],[4, 5, 6], [7, 8, 9]];
console.log(diagonalSum(matrix));

illustrate

• Initialize the variable sum to store the sum of the diagonals, and initialize the variable n to store the number of rows in the matrix.

• Use a for loop to iterate over the matrix, adding the diagonal values ​​to sum. For each iteration i, we add the main diagonal matrix[ i][i] and the antidiagonal matrix[i][n - i - 1].

• If the number of rows of the matrix is ​​odd, we subtract the middle value matrix[mid][mid] (where mid is the middle row index, use Math.floor(n / 2)) is calculated because it will be added twice.

• Returns the value of the sum.

The time complexity of this algorithm is O(n), making it an efficient solution for computing the sum of the diagonals of a matrix.

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