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, 3, 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
  1. The primary diagonal is formed by the elements A00, A11, A22, A33.

    Condition for Principal Diagonal:

    The row-column condition is row = column.
  2. 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++

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// 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++) {
        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;
}

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Java

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// 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

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Python3

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# Python3 Program to prthe Diagonals of a Matrix
MAX = 100
  
# Function to prthe 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 prthe 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

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C#

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// 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

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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 uptil 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 uptil 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 uptill i < n.

Below is the implementation of the above approach.

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// 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++) {
        // 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

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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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