Given an n x n square matrix, count all rows and columns whose sum is equal to the sum of any principal diagonal or secondary diagonal. Examples:
Input : n = 3 arr[][] = { {1, 2, 3}, {4, 5, 2}, {7, 9, 10}}; Output : 2 In first example sum of principal diagonal = (1 + 5 + 10) = 16 and sum of secondary diagonal = (3 + 5 + 7) = 15. Input: n = 4 arr[][] = { { 7, 2, 3, 5 }, { 4, 5, 6, 3 }, { 7, 9, 10, 12 }, { 1, 5, 4, 3 } }; Output: 1
We need to count number of rows or columns whose sum is equal to 16 or 15. So find sum of all rows and columns and if their sum is equal to 16 or 15 then increment the count. Therefore, sum of column {2, 5, 9} = 16 and sum column {3, 2, 10} = 15. Hence, the count is equal to 2.
C++
// C++ program to Count number of rows and // columns having sum is equal to sum of // any diagonal in matrix #include <iostream> #define n 4 using namespace std;
// function to count number of rows countnd columns // whose sum is equal to sum of any diagonal int count( int arr[][n])
{ int diag1 = 0, diag2 = 0;
int row = 0, col = 0, count = 0;
for ( int i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principal diagonal
diag1 += arr[i][i];
// sum of secondary diagonal
diag2 += arr[i][j];
}
// Find the sum of individual
// row and column
for ( int i = 0; i < n; i++) {
row = 0, col = 0;
for ( int j = 0; j < n; j++) {
row = row + arr[i][j];
}
for ( int j = 0; j < n; j++) {
col = col + arr[j][i];
}
if ((row == diag1) || (row == diag2)) {
count++;
}
if ((col == diag1) || (col == diag2))
count++;
}
return count;
} // Driver code int main()
{ int arr[n][n] = { { 7, 2, 3, 5 },
{ 4, 5, 6, 3 },
{ 7, 9, 10, 12 },
{ 1, 5, 4, 3 } };
cout << count(arr) << endl;
} |
Java
// Java program to Count number of rows and // columns having sum is equal to sum of // any diagonal in matrix import java.io.*;
class GFG {
static int n = 4 ;
// function to count number of rows countnd
// columns whose sum is equal to sum of any
// diagonal
static int count( int arr[][])
{
int diag1 = 0 , diag2 = 0 ;
int row = 0 , col = 0 , count = 0 ;
for ( int i = 0 , j = n - 1 ; i < n; i++, j--)
{
// sum of principal diagonal
diag1 += arr[i][i];
// sum of secondary diagonal
diag2 += arr[i][j];
}
// Find the sum of individual
// row and column
for ( int i = 0 ; i < n; i++) {
row = 0 ;
col = 0 ;
for ( int j = 0 ; j < n; j++) {
row = row + arr[i][j];
}
for ( int j = 0 ; j < n; j++) {
col = col + arr[j][i];
}
if ((row == diag1) || (row == diag2))
count++;
if ((col == diag1) || (col == diag2))
count++;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[][] = {{ 7 , 2 , 3 , 5 },
{ 4 , 5 , 6 , 3 },
{ 7 , 9 , 10 , 12 },
{ 1 , 5 , 4 , 3 }};
System.out.println(count(arr));
}
} // This code is contributed by vt_m. |
Python3
# Python3 program to Count number of rows # and columns having sum is equal to sum # of any diagonal in matrix n = 4
# Function to count number of rows # countnd columns whose sum is equal # to sum of any diagonal def count(arr):
diag1 = 0 ; diag2 = 0 ; row = 0
col = 0 ; count = 0 ; j = n - 1
for i in range (n):
# sum of principal diagonal
diag1 + = arr[i][i]
# sum of secondary diagonal
diag2 + = arr[i][j]
j - = 1
# Find the sum of individual
# row and column
for i in range (n):
row = 0 ; col = 0
for j in range (n):
row + = arr[i][j]
for j in range (n):
col + = arr[j][i]
if ((row = = diag1) or (row = = diag2)):
count + = 1
if ((col = = diag1) or (col = = diag2)):
count + = 1
return count
# Driver code arr = [[ 7 , 2 , 3 , 5 ],
[ 4 , 5 , 6 , 3 ],
[ 7 , 9 , 10 , 12 ],
[ 1 , 5 , 4 , 3 ] ]
print (count(arr))
# This code is contributed by Anant Agarwal. |
C#
// C# program to Count number of rows and // columns having sum is equal to sum of // any diagonal in matrix using System;
namespace Matrix
{ public class GFG
{ static int n = 4;
// Function to count number of rows
// countnd columns whose sum is equal
// to sum of any diagonal
static int count( int [,]arr)
{
int diag1 = 0, diag2 = 0;
int row = 0, col = 0, count = 0;
for ( int i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principal diagonal
diag1 += arr[i,i];
// sum of secondary diagonal
diag2 += arr[i,j];
}
// Find the sum of individual
// row and column
for ( int i = 0; i < n; i++) {
row = 0;
col = 0;
for ( int j = 0; j < n; j++) {
row = row + arr[i,j];
}
for ( int j = 0; j < n; j++) {
col = col + arr[j,i];
}
if ((row == diag1) || (row == diag2))
count++;
if ((col == diag1) || (col == diag2))
count++;
}
return count;
}
// Driver code
public static void Main()
{
int [,]arr = { { 7, 2, 3, 5 },
{ 4, 5, 6, 3 },
{ 7, 9, 10, 12 },
{ 1, 5, 4, 3 } };
Console.Write(count(arr));
}
} } // This code is contributed by Sam007 |
PHP
<?php // PHP program to Count number of rows and // columns having sum is equal to sum of // any diagonal in matrix // function to count number of // rows countnd columns whose // sum is equal to sum of any diagonal function countt( $arr )
{ $diag1 = 0; $diag2 = 0;
$row = 0; $col = 0;
$count = 0; $n = 4;
for ( $i = 0, $j = $n - 1; $i < $n ; $i ++, $j --)
{
// sum of principal diagonal
$diag1 += $arr [ $i ][ $i ];
// sum of secondary diagonal
$diag2 += $arr [ $i ][ $j ];
}
// Find the sum of individual
// row and column
for ( $i = 0; $i < $n ; $i ++) {
$row = 0; $col = 0;
for ( $j = 0; $j < $n ; $j ++) {
$row = $row + $arr [ $i ][ $j ];
}
for ( $j = 0; $j < $n ; $j ++) {
$col = $col + $arr [ $j ][ $i ];
}
if (( $row == $diag1 ) || ( $row == $diag2 )) {
$count ++;
}
if (( $col == $diag1 ) || ( $col == $diag2 ))
$count ++;
}
return $count ;
} // Driver code { $arr = array ( array (7, 2, 3, 5),
array (4, 5, 6, 3),
array (7, 9, 10, 12),
array (1, 5, 4, 3));
echo countt( $arr ) ;
} // This code is contributed by nitin mittal. ?> |
Javascript
// JS program to Count number of rows and // columns having sum is equal to sum of // any diagonal in matrix let n = 4 // function to count number of rows countnd columns // whose sum is equal to sum of any diagonal function count(arr)
{ let diag1 = 0, diag2 = 0;
let row = 0, col = 0, count = 0;
for (let i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principal diagonal
diag1 += arr[i][i];
// sum of secondary diagonal
diag2 += arr[i][j];
}
// Find the sum of individual
// row and column
for (let i = 0; i < n; i++) {
row = 0, col = 0;
for (let j = 0; j < n; j++) {
row = row + arr[i][j];
}
for (let j = 0; j < n; j++) {
col = col + arr[j][i];
}
if ((row == diag1) || (row == diag2)) {
count++;
}
if ((col == diag1) || (col == diag2))
count++;
}
return count;
} // Driver code let arr = [[ 7, 2, 3, 5 ], [ 4, 5, 6, 3 ], [ 7, 9, 10, 12 ], [ 1, 5, 4, 3]]; console.log(count(arr)) // This code is contributed by phasing17 |
Output:
1
Time complexity: O(n2) for given input n*n matrix
Auxiliary Space: O(1)