Given a 2D square matrix, find the sum of elements in Principal and Secondary diagonals. For example, consider the following 4 X 4 input matrix.
A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33
The primary diagonal is formed by the elements A00, A11, A22, A33.
- Condition for Principal Diagonal: The row-column condition is row = column.
The secondary diagonal is formed by the elements A03, A12, A21, A30.
- Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.
Input :
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output :
Principal Diagonal: 16
Secondary Diagonal: 20
Input :
3
1 1 1
1 1 1
1 1 1
Output :
Principal Diagonal: 3
Secondary Diagonal: 3
Method 1: In this method, we use two loops i.e. a loop for columns and a loop for rows and in the inner loop we check for the condition stated above:
C
#include <stdio.h>
const int M = 4;
const int N = 4;
void printDiagonalSums( int mat[M][N])
{
int principal = 0, secondary = 0;
for ( int i = 0; i < M; i++)
{
for ( int j = 0; j < N; j++)
{
if (i == j)
principal += mat[i][j];
if ((i + j) == (N - 1))
secondary += mat[i][j];
}
}
printf ( "%s" , "Principal Diagonal:" );
printf ( "%d\n" , principal);
printf ( "%s" , "Secondary Diagonal:" );
printf ( "%d\n" , secondary);
}
int main()
{
int a[][4] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printDiagonalSums(a);
return 0;
}
|
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums( int mat[][MAX], int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
if (i == j)
principal += mat[i][j];
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
cout << "Principal Diagonal:" << principal << endl;
cout << "Secondary Diagonal:" << secondary << endl;
}
int main()
{
int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
{ 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
return 0;
}
|
Java
import java.io.*;
public class GFG {
static void printDiagonalSums( int [][]mat,
int n)
{
int principal = 0 , secondary = 0 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++) {
if (i == j)
principal += mat[i][j];
if ((i + j) == (n - 1 ))
secondary += mat[i][j];
}
}
System.out.println( "Principal Diagonal:"
+ principal);
System.out.println( "Secondary Diagonal:"
+ secondary);
}
static public void main (String[] args)
{
int [][]a = { { 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 } };
printDiagonalSums(a, 4 );
}
}
|
Python3
MAX = 100
def printDiagonalSums(mat, n):
principal = 0
secondary = 0 ;
for i in range ( 0 , n):
for j in range ( 0 , n):
if (i = = j):
principal + = mat[i][j]
if ((i + j) = = (n - 1 )):
secondary + = mat[i][j]
print ( "Principal Diagonal:" , principal)
print ( "Secondary Diagonal:" , secondary)
a = [[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ],
[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ]]
printDiagonalSums(a, 4 )
|
C#
using System;
public class GFG {
static void printDiagonalSums( int [,]mat,
int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
if (i == j)
principal += mat[i,j];
if ((i + j) == (n - 1))
secondary += mat[i,j];
}
}
Console.WriteLine( "Principal Diagonal:"
+ principal);
Console.WriteLine( "Secondary Diagonal:"
+ secondary);
}
static public void Main ()
{
int [,]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
|
PHP
<?php
$MAX = 100;
function printDiagonalSums( $mat , $n )
{
global $MAX ;
$principal = 0;
$secondary = 0;
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 0; $j < $n ; $j ++)
{
if ( $i == $j )
$principal += $mat [ $i ][ $j ];
if (( $i + $j ) == ( $n - 1))
$secondary += $mat [ $i ][ $j ];
}
}
echo "Principal Diagonal:" ,
$principal , "\n" ;
echo "Secondary Diagonal:" ,
$secondary , "\n" ;
}
$a = array ( array ( 1, 2, 3, 4 ),
array ( 5, 6, 7, 8 ),
array ( 1, 2, 3, 4 ),
array ( 5, 6, 7, 8 ));
printDiagonalSums( $a , 4);
?>
|
Javascript
<script>
const MAX = 100;
void printDiagonalSums(mat, n)
{
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (i == j)
principal += mat[i][j];
if ((i + j) == (n - 1))
secondary += mat[i][j];
}
}
document.write( "Principal Diagonal:" + principal + "<br>" );
document.write( "Secondary Diagonal:" + secondary + "<br>" );
}
let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ];
printDiagonalSums(a, 4);
</script>
|
Output
Principal Diagonal:18
Secondary Diagonal:18
Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.
Method 2( Efficient Approach): In this method, we use one loop i.e. a loop for calculating the sum of both the principal and secondary diagonals:
C
#include <stdio.h>
#include <stdlib.h>
const int M = 4;
const int N = 4;
void printDiagonalSums( int mat[M][N])
{
int principal = 0, secondary = 0;
for ( int i = 0; i < N; i++)
{
principal += mat[i][i];
secondary += mat[i][N - i - 1];
}
printf ( "%s" , "Principal Diagonal:" );
printf ( "%d\n" , principal);
printf ( "%s" , "Secondary Diagonal:" );
printf ( "%d\n" , secondary);
}
int main()
{
int a[4][4] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{1, 2, 3, 4},
{5, 6, 7, 8}};
printDiagonalSums(a);
return 0;
}
|
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
void printDiagonalSums( int mat[][MAX], int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
cout << "Principal Diagonal:" << principal << endl;
cout << "Secondary Diagonal:" << secondary << endl;
}
int main()
{
int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
{ 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
return 0;
}
|
Java
import java.io.*;
public class GFG {
static void printDiagonalSums( int [][]mat,
int n)
{
int principal = 0 , secondary = 0 ;
for ( int i = 0 ; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1 ];
}
System.out.println( "Principal Diagonal:"
+ principal);
System.out.println( "Secondary Diagonal:"
+ secondary);
}
static public void main (String[] args)
{
int [][]a = { { 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 } };
printDiagonalSums(a, 4 );
}
}
|
Python3
MAX = 100
def printDiagonalSums(mat, n):
principal = 0
secondary = 0
for i in range ( 0 , n):
principal + = mat[i][i]
secondary + = mat[i][n - i - 1 ]
print ( "Principal Diagonal:" , principal)
print ( "Secondary Diagonal:" , secondary)
a = [[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ],
[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ]]
printDiagonalSums(a, 4 )
|
C#
using System;
public class GFG {
static void printDiagonalSums( int [,]mat,
int n)
{
int principal = 0, secondary = 0;
for ( int i = 0; i < n; i++) {
principal += mat[i,i];
secondary += mat[i,n - i - 1];
}
Console.WriteLine( "Principal Diagonal:"
+ principal);
Console.WriteLine( "Secondary Diagonal:"
+ secondary);
}
static public void Main ()
{
int [,]a = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 } };
printDiagonalSums(a, 4);
}
}
|
PHP
<?php
$MAX = 100;
function printDiagonalSums( $mat , $n )
{
global $MAX ;
$principal = 0; $secondary = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$principal += $mat [ $i ][ $i ];
$secondary += $mat [ $i ][ $n - $i - 1];
}
echo "Principal Diagonal:" ,
$principal , "\n" ;
echo "Secondary Diagonal:" ,
$secondary , "\n" ;
}
$a = array ( array (1, 2, 3, 4),
array (5, 6, 7, 8),
array (1, 2, 3, 4),
array (5, 6, 7, 8));
printDiagonalSums( $a , 4);
?>
|
Javascript
<script>
function printDiagonalSums(mat,n)
{
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
document.write( "Principal Diagonal:"
+ principal+ "<br>" );
document.write( "Secondary Diagonal:"
+ secondary);
}
let a = [[ 1, 2, 3, 4 ],
[5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]];
printDiagonalSums(a, 4);
</script>
|
Output
Principal Diagonal:18
Secondary Diagonal:18
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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