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 <stdio.h>#include<bits/stdc++.h>using namespace std;const int MAX = 100;// Function to Check center element // is equal to the individual // sum of all the half diagonalsbool 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 codeint 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 updatesimport 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 diagonalsdef 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 codeif __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 updatesusing 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
<?php// PHP Program to check if // the center element is // equal to the individual // sum of all the half diagonals$MAX = 100;// Function to Check center // element is equal to the // individual sum of all // the half diagonalsfunction HalfDiagonalSums($mat, $n){ global $MAX ; // Find sums of // half diagonals $diag1_left = 1; $diag1_right = 1; $diag2_left = 1; $diag2_right = 1; for ($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][$n / 2]);}// Driver code$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
<script>// Javascript Program to check if the center// element is equal to the individual // sum of all the half diagonalsconst MAX = 100;// Function to Check center element // is equal to the individual // sum of all the half diagonalsfunction HalfDiagonalSums(mat, n){ // Find sums of half diagonals let diag1_left = 0, diag1_right = 0; let diag2_left = 0, diag2_right = 0; for (let i = 0, j = n - 1; i < n; i++, j--) { if (i < parseInt(n/2)) { diag1_left += mat[i][i]; diag2_left += mat[j][i]; } else if (i > parseInt(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[parseInt(n/2)][parseInt(n/2)]);}// Driver code let 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] ]; document.write( HalfDiagonalSums(a, 5) ? "Yes" : "No" );// This code is contributed by subham348.</script> |
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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