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.
- Condition for Principal Diagonal: The row-column condition is row = column.
The secondary diagonal is formed by the elements A03, A12, A21, A30. - 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:
Java
// A simple java program to find// sum of diagonalsimport java.io.*;public class GFG { static void printDiagonalSums(int [][]mat, int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { for (int 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]; } } System.out.println("Principal Diagonal:" + principal); System.out.println("Secondary Diagonal:" + secondary); } // Driver code static public void main (String[] args) { int [][]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 vt_m. |
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:
Java
// An efficient java program to find// sum of diagonalsimport java.io.*;public class GFG { static void printDiagonalSums(int [][]mat, int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { principal += mat[i][i]; secondary += mat[i][n - i - 1]; } System.out.println("Principal Diagonal:" + principal); System.out.println("Secondary Diagonal:" + secondary); } // Driver code static public void main (String[] args) { int [][]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 vt_m. |
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!
