# Finding the converging element of the diagonals in a square matrix

• Difficulty Level : Easy
• Last Updated : 10 Aug, 2022

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.

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 ``#include ``#include ` `// 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

 ``

Output:

`6`

Time complexity: O(1) because performing constant operations

Auxiliary space: O(1)

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