Given a matrix of n X n. The task is to calculate the absolute difference between the sums of its diagonal.
Examples:
Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. Sum of primary diagonal = 4 + 5 + 10 = 19. Difference = |19 - 4| = 15. Input : mat[][] = 10 2 4 5 Output : 7
Calculate the sums across the two diagonals of a square matrix. Along the first diagonal of the matrix, row index = column index i.e mat[i][j] lies on the first diagonal if i = j. Along the other diagonal, row index = n – 1 – column index i.e mat[i][j] lies on the second diagonal if i = n-1-j. By using two loops we traverse the entire matrix and calculate the sum across the diagonals of the matrix.
Below is the implementation of this approach:
<script> // Javascript Code for Find difference between sums // of two diagonals function difference(arr,n)
{
// Initialize sums of diagonals
let d1 = 0, d2 = 0;
for (let i = 0; i < n; i++)
{
for (let j = 0; j < n; j++)
{
// finding sum of primary diagonal
if (i == j)
d1 += arr[i][j];
// finding sum of secondary diagonal
if (i == n - j - 1)
d2 += arr[i][j];
}
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
let n = 3;
let arr =
[
[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]
];
document.write(difference(arr, n));
// This code is contributed Bobby </script> |
Output:
15
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.
We can optimize above solution to work in O(n) using the patterns present in indexes of cells.
<script> // JAVA SCRIPT Code for Find difference between sums // of two diagonals function difference(arr,n)
{
// Initialize sums of diagonals
let d1 = 0, d2 = 0;
for (let i = 0; i < n; i++)
{
d1 += arr[i][i];
d2 += arr[i][n-i-1];
}
// Absolute difference of the sums
// across the diagonals
return Math.abs(d1 - d2);
}
/* Driver program to test above function */
let n = 3;
let arr = [[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]];
document.write(difference(arr, n));
// This code is contributed by sravan kumar Gottumukkala </script> |
Output:
15
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 Find difference between sums of two diagonals for more details!