Program to print the Diagonals of a Matrix in O(N) time

Given a 2D square matrix, the task is to print the Principal and Secondary diagonals of this matrix in O(N) time complexity. For O(N²) time, please refer this article. 

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

Approach:

1.Consider the following 4 X 4 input matrix. 

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

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

Condition for Principal Diagonal: 

The row-column condition is row = column.

3.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.

In this method, we use one loop i.e. a loop to find the diagonal elements as per the below formula:

principal diagonal = matrix[i][i];
secondary diagonal = matrix[i][n - i - 1];

where 0 <= i <= n

Below is the implementation of the above approach: 

C++

// C++ Program to print the Diagonals of a Matrix

#include <bits/stdc++.h>
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++) {

        // Condition for principal diagonal
        cout << mat[i][i] << ", ";
    }
    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++) {

        // Condition for secondary diagonal
        cout << mat[i][n - i - 1] << ", ";
    }

    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

// Java Program to print the Diagonals of a Matrix
class GFG
{

    static final 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++)
        {

            // Condition for principal diagonal
            System.out.print(mat[i][i] + ", ");
        }
        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++)
        {

            // Condition for secondary diagonal
            System.out.print(mat[i][n - i - 1] + ", ");
        }

        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 AnkitRai01

Python3

# Python 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):

        # Condition for principal diagonal
        print(mat[i][i], end= ", ");

    print();

# Function to print the Secondary Diagonal
def printSecondaryDiagonal(mat, n):
    print("Secondary Diagonal: ", end = "");

    for i in range(n):

        # Condition for secondary diagonal
        print(mat[i][n - i - 1], end = ", ");

    print();

# Driver code
if __name__ == '__main__':
    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 PrinciRaj1992

C#

// C# Program to print the Diagonals of a Matrix
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++)
        {

            // Condition for principal diagonal
            Console.Write(mat[i, i] + ", ");
        }
        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++)
        {

            // Condition for secondary diagonal
            Console.Write(mat[i, n - i - 1] + ", ");
        }

        Console.WriteLine();
    }

    // Driver code
    public 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 AnkitRai01

Javascript

<script>
// Java script Program to print the Diagonals of a Matrix
let MAX = 100;

    // Function to print the Principal Diagonal
    function printPrincipalDiagonal(mat,n)
    {
        document.write("Principal Diagonal: ");

        for (let i = 0; i < n; i++)
        {

            // Condition for principal diagonal
            document.write(mat[i][i] + ", ");
        }
        document.write("<br>");
    }

    // Function to print the Secondary Diagonal
    function printSecondaryDiagonal(mat,n)
    {
        document.write("Secondary Diagonal: ");

        for (let i = 0; i < n; i++)
        {

            // Condition for secondary diagonal
            document.write(mat[i][n - i - 1] + ", ");
        }

        document.write("<br>");
    }

    // Driver code

        let n = 4;
        let 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 sravan kumar Gottumukklala
</script>

Output:

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

Time complexity: O(n) for given n

Auxiliary space: O(1) as it is using constant space