# Program to print the Diagonals of a Matrix

Given a 2D square matrix, print the Principal and Secondary diagonals.
Examples :

```Input:
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output:
Principal Diagonal: 1, 3, 9, 3
Secondary Diagonal: 4, 2, 8, 6

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

```

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.
```

Method 1:
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.

 `// C++ Program to print the Diagonals of a Matrix`   `#include ` `using` `namespace` `std;`   `const` `int` `MAX = 100;`   `// Function to print the Principal Diagonal` `void` `printPrincipalDiagonal(``int` `mat[][MAX], ``int` `n)` `{` `    ``cout << ``"Principal Diagonal: "``;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = 0; j < n; j++) {`   `            ``// Condition for principal diagonal` `            ``if` `(i == j)` `                ``cout << mat[i][j] << ``", "``;` `        ``}` `    ``}` `    ``cout << endl;` `}`   `// Function to print the Secondary Diagonal` `void` `printSecondaryDiagonal(``int` `mat[][MAX], ``int` `n)` `{` `    ``cout << ``"Secondary Diagonal: "``;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = 0; j < n; j++) {`   `            ``// Condition for secondary diagonal` `            ``if` `((i + j) == (n - 1))` `                ``cout << mat[i][j] << ``", "``;` `        ``}` `    ``}` `    ``cout << endl;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 4;` `    ``int` `a[][MAX] = { { 1, 2, 3, 4 },` `                     ``{ 5, 6, 7, 8 },` `                     ``{ 1, 2, 3, 4 },` `                     ``{ 5, 6, 7, 8 } };`   `    ``printPrincipalDiagonal(a, n);` `    ``printSecondaryDiagonal(a, n);` `    ``return` `0;` `}`

 `// Java Program to print the Diagonals of a Matrix` `class` `GFG {` `    ``static` `int` `MAX = ``100``;`   `    ``// Function to print the Principal Diagonal` `    ``static` `void` `printPrincipalDiagonal(``int` `mat[][], ``int` `n)` `    ``{` `        ``System.out.print(``"Principal Diagonal: "``);`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``for` `(``int` `j = ``0``; j < n; j++) {`   `                ``// Condition for principal diagonal` `                ``if` `(i == j) {` `                    ``System.out.print(mat[i][j] + ``", "``);` `                ``}` `            ``}` `        ``}` `        ``System.out.println(``""``);` `    ``}`   `    ``// Function to print the Secondary Diagonal` `    ``static` `void` `printSecondaryDiagonal(``int` `mat[][], ``int` `n)` `    ``{` `        ``System.out.print(``"Secondary Diagonal: "``);`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``for` `(``int` `j = ``0``; j < n; j++) {`   `                ``// Condition for secondary diagonal` `                ``if` `((i + j) == (n - ``1``)) {` `                    ``System.out.print(mat[i][j] + ``", "``);` `                ``}` `            ``}` `        ``}` `        ``System.out.println(``""``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `n = ``4``;` `        ``int` `a[][] = { { ``1``, ``2``, ``3``, ``4` `},` `                      ``{ ``5``, ``6``, ``7``, ``8` `},` `                      ``{ ``1``, ``2``, ``3``, ``4` `},` `                      ``{ ``5``, ``6``, ``7``, ``8` `} };`   `        ``printPrincipalDiagonal(a, n);` `        ``printSecondaryDiagonal(a, n);` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

 `# Python3 Program to print the Diagonals of a Matrix` `MAX` `=` `100`   `# Function to print the Principal Diagonal` `def` `printPrincipalDiagonal(mat, n):` `    ``print``(``"Principal Diagonal: "``, end ``=` `"")`   `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(n):`   `            ``# Condition for principal diagonal` `            ``if` `(i ``=``=` `j):` `                ``print``(mat[i][j], end ``=` `", "``)` `    ``print``()`   `# Function to print the Secondary Diagonal` `def` `printSecondaryDiagonal(mat, n):` `    ``print``(``"Secondary Diagonal: "``, end ``=` `"")`   `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(n):`   `            ``# Condition for secondary diagonal` `            ``if` `((i ``+` `j) ``=``=` `(n ``-` `1``)):` `                ``print``(mat[i][j], end ``=` `", "``)` `    ``print``()`   `# Driver code` `n ``=` `4` `a ``=` `[[ ``1``, ``2``, ``3``, ``4` `],` `     ``[ ``5``, ``6``, ``7``, ``8` `],` `     ``[ ``1``, ``2``, ``3``, ``4` `],` `     ``[ ``5``, ``6``, ``7``, ``8` `]]`   `printPrincipalDiagonal(a, n)` `printSecondaryDiagonal(a, n)`   `# This code is contributed by Mohit Kumar`

 `// C# Program to print the Diagonals of a Matrix` `using` `System;`   `class` `GFG {` `    ``static` `int` `MAX = 100;`   `    ``// Function to print the Principal Diagonal` `    ``static` `void` `printPrincipalDiagonal(``int``[, ] mat, ``int` `n)` `    ``{` `        ``Console.Write(``"Principal Diagonal: "``);`   `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``for` `(``int` `j = 0; j < n; j++) {`   `                ``// Condition for principal diagonal` `                ``if` `(i == j) {` `                    ``Console.Write(mat[i, j] + ``", "``);` `                ``}` `            ``}` `        ``}` `        ``Console.WriteLine(``""``);` `    ``}`   `    ``// Function to print the Secondary Diagonal` `    ``static` `void` `printSecondaryDiagonal(``int``[, ] mat, ``int` `n)` `    ``{` `        ``Console.Write(``"Secondary Diagonal: "``);`   `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``for` `(``int` `j = 0; j < n; j++) {`   `                ``// Condition for secondary diagonal` `                ``if` `((i + j) == (n - 1)) {` `                    ``Console.Write(mat[i, j] + ``", "``);` `                ``}` `            ``}` `        ``}` `        ``Console.WriteLine(``""``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int` `n = 4;` `        ``int``[, ] a = { { 1, 2, 3, 4 },` `                      ``{ 5, 6, 7, 8 },` `                      ``{ 1, 2, 3, 4 },` `                      ``{ 5, 6, 7, 8 } };`   `        ``printPrincipalDiagonal(a, n);` `        ``printSecondaryDiagonal(a, n);` `    ``}` `}`   `// This code is contributed by 29AjayKumar`

Output:
```Principal Diagonal: 1, 6, 3, 8,
Secondary Diagonal: 4, 7, 2, 5,

```

Complexity Analysis:

• Time Complexity: O(n2).
As there is a nested loop involved so the time complexity is squared.
• Auxiliary Space: O(1).
As no extra space is occupied.

Method 2:
In this method, the same condition for printing the diagonal elements can be achieved using a single for loop.
Approach:

1. For Principal Diagonal elements: Run a for loop until n, where n is the number of columns, and print array[i][i] where i is the index variable.
2. For Secondary Diagonal elements: Run a for loop until n, where n is the number of columns and print array[i][k] where i is the index variable and k = array_length – 1. Decrease k until i < n

Below is the implementation of the above approach.

 `// C++ Program to print the Diagonals of a Matrix`   `#include ` `using` `namespace` `std;`   `const` `int` `MAX = 100;`   `// Function to print the Principal Diagonal` `void` `printPrincipalDiagonal(``int` `mat[][MAX], ``int` `n)` `{` `    ``cout << ``"Principal Diagonal: "``;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Printing principal diagonal` `        ``cout << mat[i][i] << ``", "``;` `    ``}` `    ``cout << endl;` `}`   `// Function to print the Secondary Diagonal` `void` `printSecondaryDiagonal(``int` `mat[][MAX], ``int` `n)` `{` `    ``cout << ``"Secondary Diagonal: "``;` `    ``int` `k = n - 1;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Printing secondary diagonal` `        ``cout << mat[i][k--] << ``", "``;` `    ``}` `    ``cout << endl;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 4;` `    ``int` `a[][MAX] = { { 1, 2, 3, 4 },` `                     ``{ 5, 6, 7, 8 },` `                     ``{ 1, 2, 3, 4 },` `                     ``{ 5, 6, 7, 8 } };`   `    ``printPrincipalDiagonal(a, n);` `    ``printSecondaryDiagonal(a, n);` `    ``return` `0;` `}`   `// This code is contributed by yashbeersingh42`

 `// C# program for the ` `// above approach` `using` `System;` `class` `GFG{` `    `  `// Function to print the ` `// Principal Diagonal` `static` `void` `printPrincipalDiagonal(``int` `[,]mat, ` `                                   ``int` `n)` `{` `  ``Console.Write(``"Principal Diagonal: "``);`   `  ``for` `(``int` `i = 0; i < n; i++) ` `  ``{` `    ``// Printing principal diagonal` `    ``Console.Write(mat[i, i] + ``", "``);` `  ``}` `  ``Console.Write(``"\n"``);` `}` ` `  `// Function to print the ` `// Secondary Diagonal` `static` `void` `printSecondaryDiagonal(``int` `[,]mat, ` `                                   ``int` `n)` `{` `  ``Console.Write(``"Secondary Diagonal: "``);` `  ``int` `k = n - 1;` `  `  `  ``for` `(``int` `i = 0; i < n; i++) ` `  ``{` `    ``// Printing secondary diagonal` `    ``Console.Write(mat[i, k--] + ``", "``);` `  ``}` `  `  `  ``Console.Write(``"\n"``);` `}` `    `  `    `  `// Driver code` `static` `void` `Main() ` `{` `  ``int` `n = 4;` `  ``int` `[,]a = {{1, 2, 3, 4},` `              ``{5, 6, 7, 8},` `              ``{1, 2, 3, 4},` `              ``{5, 6, 7, 8}};` `  ``printPrincipalDiagonal(a, n);` `  ``printSecondaryDiagonal(a, n);` `}` `}`   `// This code is contributed by rutvik_56`

Output:
```Principal Diagonal: 1, 6, 3, 8,
Secondary Diagonal: 4, 7, 2, 5,

```

Complexity Analysis:

• Time Complexity: O(n).
As there is only one loop involved so the time complexity is linear.
• Auxiliary Space: O(1).
As no extra space is occupied.

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