# Center element of matrix equals sums of half diagonals

• Difficulty Level : Basic
• Last Updated : 09 Aug, 2022

Given a matrix of odd order i.e(5*5). Task is to check if the center element of the matrix is equal to the individual sum of all the half diagonals.

Examples:

```Input : mat[][] = {   2   9   1   4  -2
6   7   2  11   4
4    2  9   2   4
1   9   2    4  4
0   2   4    2  5 }
Output : Yes
Explanation :
Sum of Half Diagonal 1 = 2 + 7 = 9
Sum of Half Diagonal 2 = 9 + 0 = 9
Sum of Half Diagonal 3 = 11 + -2 = 9
Sum of Half Diagonal 4 = 5 + 4 = 9
Here, All the sums equal to the center element
that is mat[2][2] = 9```

Simple Approach:
Iterate two loops, find all half diagonal sums and then check all sums are equal to the center element of the matrix or not. If any one of them is not equal to center element Then print “No” Else “Yes”.
Time Complexity: O(N*N)

Efficient Approach : is based on Efficient approach to find diagonal sum in O(N)

Below are the Implementation of this approach

## C++

 `// C++ Program to check if the center``// element is equal to the individual``// sum of all the half diagonals``#include ``#include``using` `namespace` `std;` `const` `int` `MAX = 100;` `// Function to Check center element``// is equal to the individual``// sum of all the half diagonals``bool` `HalfDiagonalSums(``int` `mat[][MAX], ``int` `n)``{   ``    ``// Find sums of half diagonals``    ``int` `diag1_left = 0, diag1_right = 0;``    ``int` `diag2_left = 0, diag2_right = 0;   ``    ``for` `(``int` `i = 0, j = n - 1; i < n; i++, j--) {``        ` `        ``if` `(i < n/2) {``            ``diag1_left += mat[i][i];``            ``diag2_left += mat[j][i];          ``        ``}``        ``else` `if` `(i > n/2) {``            ``diag1_right += mat[i][i];``            ``diag2_right += mat[j][i];          ``        ``}``    ``}``    ` `    ``return` `(diag1_left == diag2_right &&``            ``diag2_right == diag2_left &&``            ``diag1_right == diag2_left &&``            ``diag2_right == mat[n/2][n/2]);``}` `// Driver code``int` `main()``{``    ``int` `a[][MAX] = { { 2, 9, 1, 4, -2},``                     ``{ 6, 7, 2, 11, 4},``                     ``{ 4, 2, 9, 2, 4},``                     ``{ 1, 9, 2, 4, 4},``                     ``{ 0, 2, 4, 2, 5} };``        ``cout << ( HalfDiagonalSums(a, 5) ? ``"Yes"` `: ``"No"` `);``    ``return` `0;``}`

## Java

 `// Java program to find maximum elements``// that can be made equal with k updates``import` `java.util.Arrays;``public` `class` `GFG {``    ` `    ``static` `int` `MAX = ``100``;``    ` `    ``// Function to Check center element``    ``// is equal to the individual``    ``// sum of all the half diagonals``    ``static` `boolean` `HalfDiagonalSums(``int` `mat[][],``                                          ``int` `n)``    ``{``        ` `        ``// Find sums of half diagonals``        ``int` `diag1_left = ``0``, diag1_right = ``0``;``        ``int` `diag2_left = ``0``, diag2_right = ``0``;``        ``for` `(``int` `i = ``0``, j = n - ``1``; i < n;``                                    ``i++, j--)``        ``{``            ` `            ``if` `(i < n/``2``) {``                ``diag1_left += mat[i][i];``                ``diag2_left += mat[j][i];        ``            ``}``            ``else` `if` `(i > n/``2``) {``                ``diag1_right += mat[i][i];``                ``diag2_right += mat[j][i];        ``            ``}``        ``}``        ` `        ``return` `(diag1_left == diag2_right &&``                ``diag2_right == diag2_left &&``                ``diag1_right == diag2_left &&``                ``diag2_right == mat[n/``2``][n/``2``]);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ` `        ``int` `a[][] = { { ``2``, ``9``, ``1``, ``4``, -``2``},``                      ``{ ``6``, ``7``, ``2``, ``11``, ``4``},``                      ``{ ``4``, ``2``, ``9``, ``2``, ``4``},``                      ``{ ``1``, ``9``, ``2``, ``4``, ``4``},``                      ``{ ``0``, ``2``, ``4``, ``2``, ``5``} };``                      ` `        ``System.out.print ( HalfDiagonalSums(a, ``5``)``                                ``? ``"Yes"` `: ``"No"` `);``    ``}``}` `// This code is contributed by Sam007`

## Python 3

 `# Python 3 Program to check if the center``# element is equal to the individual``# sum of all the half diagonals`` ` `MAX` `=` `100`` ` `# Function to Check center element``# is equal to the individual``# sum of all the half diagonals``def` `HalfDiagonalSums( mat,  n):` `    ``# Find sums of half diagonals``    ``diag1_left ``=` `0``    ``diag1_right ``=` `0``    ``diag2_left ``=` `0``    ``diag2_right ``=` `0` `    ``i ``=` `0``    ``j ``=` `n ``-` `1``    ``while` `i < n:``         ` `        ``if` `(i < n``/``/``2``) :``            ``diag1_left ``+``=` `mat[i][i]``            ``diag2_left ``+``=` `mat[j][i]          ``        ` `        ``elif` `(i > n``/``/``2``) :``            ``diag1_right ``+``=` `mat[i][i]``            ``diag2_right ``+``=` `mat[j][i]          ``        ``i ``+``=` `1``        ``j ``-``=` `1``     ` `    ``return` `(diag1_left ``=``=` `diag2_right ``and``            ``diag2_right ``=``=` `diag2_left ``and``            ``diag1_right ``=``=` `diag2_left ``and``            ``diag2_right ``=``=` `mat[n``/``/``2``][n``/``/``2``])`` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``a ``=` `[[``2``, ``9``, ``1``, ``4``, ``-``2``],``         ``[``6``, ``7``, ``2``, ``11``, ``4``],``         ``[ ``4``, ``2``, ``9``, ``2``, ``4``],``         ``[``1``, ``9``, ``2``, ``4``, ``4` `],``         ``[ ``0``, ``2``, ``4``, ``2``, ``5``]]``    ` `    ``print``(``"Yes"``) ``if` `(HalfDiagonalSums(a, ``5``)) ``else` `print``(``"No"` `)`

## C#

 `// C# program to find maximum``// elements that can be made``// equal with k updates``using` `System;` `class` `GFG``{` `    ``// Function to Check``    ``// center element is``    ``// equal to the individual``    ``// sum of all the half``    ``// diagonals``    ``static` `bool` `HalfDiagonalSums(``int` `[,]mat,``                                 ``int` `n)``    ``{``        ` `        ``// Find sums of``        ``// half diagonals``        ``int` `diag1_left = 0,``            ``diag1_right = 0;``        ``int` `diag2_left = 0,``            ``diag2_right = 0;``        ``for` `(``int` `i = 0, j = n - 1;``                 ``i < n; i++, j--)``        ``{``            ` `            ``if` `(i < n / 2)``            ``{``                ``diag1_left += mat[i, i];``                ``diag2_left += mat[j, i];    ``            ``}``            ``else` `if` `(i > n / 2)``            ``{``                ``diag1_right += mat[i, i];``                ``diag2_right += mat[j, i];        ``            ``}``        ``}``        ` `        ``return` `(diag1_left == diag2_right &&``                ``diag2_right == diag2_left &&``                ``diag1_right == diag2_left &&``                ``diag2_right == mat[n / 2, n / 2]);``    ``}``    ` `    ``// Driver code``    ``static` `public` `void` `Main ()``    ``{``        ``int` `[,]a = {{ 2, 9, 1, 4, -2},``                    ``{ 6, 7, 2, 11, 4},``                    ``{ 4, 2, 9, 2, 4},``                    ``{ 1, 9, 2, 4, 4},``                    ``{ 0, 2, 4, 2, 5}};``                    ` `        ``Console.WriteLine(HalfDiagonalSums(a, 5)?``                                  ``"Yes"` `: ``"No"` `);``    ``}``}` `// This code is contributed by ajit`

## PHP

 ` ``\$n` `/ 2)``        ``{``            ``\$diag1_right` `+= ``\$mat``[``\$i``][``\$i``];``            ``\$diag2_right` `+= ``\$mat``[``\$j``][``\$i``];    ``        ``}``    ``}``    ` `    ``return` `(``\$diag1_left` `== ``\$diag2_right` `&&``            ``\$diag2_right` `== ``\$diag2_left` `&&``            ``\$diag1_right` `== ``\$diag2_left` `&&``            ``\$diag2_right` `== ``\$mat``[``\$n` `/ 2][``\$n` `/ 2]);``}` `// Driver code``\$a` `= ``array``(``array``(2, 9, 1, 4, -2),``           ``array``(6, 7, 2, 11, 4),``           ``array``(4, 2, 9, 2, 4),``           ``array``(1, 9, 2, 4, 4),``           ``array``(0, 2, 4, 2, 5));``if``(HalfDiagonalSums(``\$a``, 5) == 0)``    ``echo` `"Yes"` `;``else``    ``echo` `"No"` `;``        ` `// This code is contributed``// by akt_mit``?>`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.

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