Center element of matrix equals sums of half diagonals
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++
#include <stdio.h>
#include<bits/stdc++.h>
using namespace std;
const int MAX = 100;
bool HalfDiagonalSums(int mat[][MAX], int n)
{
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]);
}
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
import java.util.Arrays;
public class GFG {
static int MAX = 100;
static boolean HalfDiagonalSums(int mat[][],
int n)
{
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]);
}
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" );
}
}
|
Python 3
MAX = 100
def HalfDiagonalSums( mat, n):
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])
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#
using System;
class GFG
{
static bool HalfDiagonalSums(int [,]mat,
int n)
{
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]);
}
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" );
}
}
|
PHP
<?php
$MAX = 100;
function HalfDiagonalSums($mat, $n)
{
global $MAX ;
$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]);
}
$a = array(array(2, 9, 1, 4, -2),
array(6, 7, 2, 11, 4),
array(4, 2, 9, 2, 4),
array(1, 9, 2, 4, 4),
array(0, 2, 4, 2, 5));
if(HalfDiagonalSums($a, 5) == 0)
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
const MAX = 100;
function HalfDiagonalSums(mat, n)
{
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)]);
}
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" );
</script>
|
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.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...