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Count rows/columns with sum equals to diagonal sum

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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)



Last Updated : 20 Feb, 2023
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