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Javascript Program to Efficiently compute sums of diagonals of a matrix

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Given a 2D square matrix, find the sum of elements in Principal and Secondary diagonals. For example, consider the following 4 X 4 input matrix.
 

A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33

The primary diagonal is formed by the elements A00, A11, A22, A33. 
 

  1. Condition for Principal Diagonal: The row-column condition is row = column. 
    The secondary diagonal is formed by the elements A03, A12, A21, A30.
  2. Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.

Examples : 
 

Input : 
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output :
Principal Diagonal: 16
Secondary Diagonal: 20

Input :
3
1 1 1
1 1 1
1 1 1
Output :
Principal Diagonal: 3
Secondary Diagonal: 3

 

 

Method 1 (O(n ^ 2) :

In this method, we use two loops i.e. a loop for columns and a loop for rows and in the inner loop we check for the condition stated above:
 

Javascript




<script>
// A simple Javascript program to find sum of diagonals
 
const MAX = 100;
 
void printDiagonalSums(mat, n)
{
    let principal = 0, secondary = 0;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
 
            // Condition for principal diagonal
            if (i == j)
                principal += mat[i][j];
 
            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
                secondary += mat[i][j];
        }
    }
 
    document.write("Principal Diagonal:" + principal + "<br>");
    document.write("Secondary Diagonal:" + secondary + "<br>");
}
 
// Driver code
    let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ];
    printDiagonalSums(a, 4);
 
// This code is contributed by subhammahato348.
</script>

Output:  

Principal Diagonal:18
Secondary Diagonal:18

Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.

Auxiliary Space: O(1), as we are not using any extra space.
 

Method 2 (O(n) :

In this method we use one loop i.e. a loop for calculating sum of both the principal and secondary diagonals: 
 

Javascript




<script>
 
// An efficient Javascript  program to find
// sum of diagonals
 
function  printDiagonalSums(mat,n)
    {
        let principal = 0, secondary = 0;
        for (let i = 0; i < n; i++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1];
        }
     
        document.write("Principal Diagonal:"
                                + principal+"<br>");
                                     
        document.write("Secondary Diagonal:"
                                + secondary);
    }
     
    // Driver code
     
        let a = [[ 1, 2, 3, 4 ],
                    [5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ]];
     
        printDiagonalSums(a, 4);
         
// This code is contributed Bobby
 
</script>

Output :  

Principal Diagonal:18
Secondary Diagonal:18

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Efficiently compute sums of diagonals of a matrix for more details!


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Last Updated : 31 May, 2022
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