Finding the converging element of the diagonals in a square matrix

Given a square matrix, the task is to find the element of the matrix where the right and the left diagonal of this square matrix converge.
Example:

Input: n = 5, matrix = 
[ 1 2 3 4 5
  5 6 7 8 6
  9 5 6 8 7
  2 3 5 6 8
  1 2 3 4 5 ]
Output: 6

Input: n = 4, matrix = 
[ 1 2 3 4
  5 6 7 8
  9 0 1 2 
  4 5 6 1 ]
Output: NULL
Here there no converging element at all. 
Hence the answer is null.

Approach:

• If the number of rows and column of the matrix are even, then we just print NULL because there would be no converging element in the case of even number of rows and column.
• If the number of rows and column of the matrix are odd, find the mid-value of n as 
  

mid = n/2

• The arr[mid][mid] itself is the converging diagonal element.

Below is the implementation of the above approach:

C++

// C++ program to find the converging element
// of the diagonals in a square matrix

#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>

// Driver code
int main()
{
    int n = 5;
    int a[][5] = { { 1, 2, 3, 4, 5 },
                   { 5, 6, 7, 8, 6 },
                   { 9, 5, 6, 8, 7 },
                   { 2, 3, 5, 6, 8 },
                   { 1, 2, 3, 4, 5 } };

    int convergingele, mid;
    int i, j;

    // If n is even, then convergence
    // element will be null.
    if (n % 2 == 0) {
        printf("NULL\n");
    }

    else {
        // finding the mid
        mid = n / 2;

        // finding the converging element
        convergingele = a[mid][mid];

        printf("%d\n", convergingele);
    }
}

Java

// Java program to find the converging element
// of the diagonals in a square matrix
class GFG
{

    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        int a[][] = {{1, 2, 3, 4, 5},
                     {5, 6, 7, 8, 6},
                     {9, 5, 6, 8, 7},
                     {2, 3, 5, 6, 8},
                     {1, 2, 3, 4, 5}};

        int convergingele, mid;
        int i, j;

        // If n is even, then convergence
        // element will be null.
        if (n % 2 == 0)
        {
            System.out.printf("NULL\n");
        }
        else
        {
            // finding the mid
            mid = n / 2;

            // finding the converging element
            convergingele = a[mid][mid];

            System.out.printf("%d\n", convergingele);
        }
    }
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the converging element
# of the diagonals in a square matrix

# Driver code
n = 5
a = [[ 1, 2, 3, 4, 5 ],
     [ 5, 6, 7, 8, 6 ],
     [ 9, 5, 6, 8, 7 ],
     [ 2, 3, 5, 6, 8 ],
     [ 1, 2, 3, 4, 5 ]]

# If n is even, then convergence
# element will be null.
if (n % 2 == 0):
    print("NULL")
else :

    # finding the mid
    mid = n // 2

    # finding the converging element
    convergingele = a[mid][mid]

    print(convergingele)

# This code is contributed by Mohit Kumar

C#

// C# program to find the converging element
// of the diagonals in a square matrix
using System;

class GFG
{

    // Driver code
    public static void Main(String []args)
    {
        int n = 5;
        int [,]a = {{1, 2, 3, 4, 5},
                    {5, 6, 7, 8, 6},
                    {9, 5, 6, 8, 7},
                    {2, 3, 5, 6, 8},
                    {1, 2, 3, 4, 5}};

        int convergingele, mid;

        // If n is even, then convergence
        // element will be null.
        if (n % 2 == 0)
        {
            Console.Write("NULL\n");
        }
        else
        {
            // finding the mid
            mid = n / 2;

            // finding the converging element
            convergingele = a[mid,mid];

            Console.Write("{0}\n", convergingele);
        }
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find the converging element
// of the diagonals in a square matrix

// Driver code
    let n = 5;
    let a = [ [ 1, 2, 3, 4, 5 ],
                   [ 5, 6, 7, 8, 6 ],
                   [ 9, 5, 6, 8, 7 ],
                   [ 2, 3, 5, 6, 8 ],
                   [ 1, 2, 3, 4, 5 ] ];

    let convergingele, mid;
    let i, j;

    // If n is even, then convergence
    // element will be null.
    if (n % 2 == 0) {
        document.write("NULL<br>");
    }

    else
    {

        // finding the mid
        mid = parseInt(n / 2);

        // finding the converging element
        convergingele = a[mid][mid];

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

// This code is contributed by subhammahato348.
</script>

Output:

6

Time complexity: O(1) because performing constant operations

Auxiliary space: O(1)