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.Â
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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:Â
1.Consider the following 4 X 4 input matrix.Â
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.
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Condition for Principal Diagonal:Â
The row-column condition is row = column.
3.The secondary diagonal is formed by the elements A03, A12, A21, A30.Â
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Condition for Secondary Diagonal:Â
The row-column condition is row = numberOfRows - column - 1.
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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:Â
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C++
// C++ Program to print the Diagonals of a MatrixÂ
#include <bits/stdc++.h>using namespace std;Â
const int MAX = 100;Â
// Function to print the Principal Diagonalvoid printPrincipalDiagonal(int mat[][MAX], int n){Â Â Â Â cout << "Principal Diagonal: ";Â
    for (int i = 0; i < n; i++) {Â
        // Condition for principal diagonal        cout << mat[i][i] << ", ";    }    cout << endl;}Â
// Function to print the Secondary Diagonalvoid printSecondaryDiagonal(int mat[][MAX], int n){Â Â Â Â cout << "Secondary Diagonal: ";Â
    for (int i = 0; i < n; i++) {Â
        // Condition for secondary diagonal        cout << mat[i][n - i - 1] << ", ";    }Â
    cout << endl;}Â
// Driver codeint main(){    int n = 4;    int a[][MAX] = { { 1, 2, 3, 4 },                     { 5, 6, 7, 8 },                     { 1, 2, 3, 4 },                     { 5, 6, 7, 8 } };Â
    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);     } }Â
// This code is contributed by AnkitRai01 |
Python3
# Python Program to print the Diagonals of a Matrix MAX = 100;Â
# Function to print the Principal Diagonaldef printPrincipalDiagonal(mat, n):Â Â Â Â print("Principal Diagonal: ", end = "");Â
    for i in range(n):Â
        # Condition for principal diagonal        print(mat[i][i], end= ", ");         print();Â
# Function to print the Secondary Diagonaldef printSecondaryDiagonal(mat, n):Â Â Â Â print("Secondary Diagonal: ", end = "");Â
    for i in range(n):Â
        # Condition for secondary diagonal        print(mat[i][n - i - 1], end = ", ");         print();Â
# Driver codeif __name__ == '__main__':    n = 4;    a = [[ 1, 2, 3, 4 ],        [ 5, 6, 7, 8 ],        [ 1, 2, 3, 4 ],        [ 5, 6, 7, 8 ]];Â
    printPrincipalDiagonal(a, n);    printSecondaryDiagonal(a, n);Â
# This code is contributed by PrinciRaj1992 |
C#
// C# Program to print the Diagonals of a Matrix using System;Â
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);     } } Â
// This code is contributed by AnkitRai01 |
Javascript
<script>// Java script Program to print the Diagonals of a Matrixlet 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);Â
// This code is contributed by sravan kumar Gottumukklala</script> |
Output:Â
Principal Diagonal: 1, 6, 3, 8, Secondary Diagonal: 4, 7, 2, 5,
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Time complexity: O(n) for given n
Auxiliary space: O(1) as it is using constant space
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