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Sum of matrix in which each element is absolute difference of its row and column numbers

Given a positive integer n. Consider a matrix of n rows and n columns, in which each element contain absolute difference of its row number and numbers. The task is to calculate sum of each element of the matrix.

Examples :

Input : n = 2
Output : 2
Matrix formed with n = 2 with given constraint:
0 1
1 0
Sum of matrix = 2.

Input : n = 3
Output : 8
Matrix formed with n = 3 with given constraint:
0 1 2
1 0 1
2 1 0
Sum of matrix = 8.



Method 1 (Brute Force):
Simply construct a matrix of n rows and n columns and initialize each cell with absolute difference of its corresponding row number and column number. Now, find the sum of each cell.

Below is the implementation of above idea :

C++

// C++ program to find sum of matrix in which each
// element is absolute difference of its corresponding
// row and column number row.
#include<bits/stdc++.h>
using namespace std;
  
// Retuen the sum of matrix in which each element
// is absolute difference of its corresponding row
// and column number row
int findSum(int n)
{
    // Generate matrix
    int arr[n][n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i][j] = abs(i - j);
  
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i][j];
  
    return sum;
}
  
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix
// in which each element is absolute
// difference of its corresponding
// row and column number row.
import java.io.*;
  
public class GFG {
  
// Retuen the sum of matrix in which
// each element is absolute difference
// of its corresponding row and column
// number row
static int findSum(int n)
{
      
    // Generate matrix
    int [][]arr = new int[n][n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i][j] = Math.abs(i - j);
  
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i][j];
  
    return sum;
}
  
    // Driver Code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
  
// This code is contributed by vt_m.

Python3

# Python3 program to find sum of matrix 
# in which each element is absolute 
# difference of its corresponding
# row and column number row.
  
# Return the sum of matrix in which each 
# element is absolute difference of its 
# corresponding row and column number row
def findSum(n):
  
    # Generate matrix
    arr = [[0 for x in range(n)]
              for y in range (n)]
    for i in range (n):
        for j in range (n):
            arr[i][j] = abs(i - j)
  
    # Compute sum
    sum = 0
    for i in range (n):
        for j in range(n):
            sum += arr[i][j]
  
    return sum
  
# Driver Code
if __name__ == "__main__":
  
    n = 3
    print (findSum(n))
      
# This code is contributed by ita_c

C#

// C# program to find sum of matrix
// in which each element is absolute
// difference of its corresponding
// row and column number row.
using System;
  
public class GFG {
  
// Retuen the sum of matrix in which
// each element is absolute difference
// of its corresponding row and column
// number row
static int findSum(int n)
{
      
// Generate matrix
    int [,]arr = new int[n, n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            arr[i,j ] = Math.Abs(i - j);
   
    // Compute sum
    int sum = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += arr[i, j];
   
    return sum;
}
  
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum of 
// matrix in which each element
// is absolute difference of 
// its corresponding row and 
// column number row.
  
// Retuen the sum of matrix 
// in which each element
// is absolute difference 
// of its corresponding row
// and column number row
function findSum( $n)
{
      
    // Generate matrix
    $arr =array(array());
    for($i = 0; $i < $n; $i++)
        for($j = 0; $j < $n; $j++)
            $arr[$i][$j] = abs($i - $j);
  
    // Compute sum
    $sum = 0;
    for($i = 0; $i < $n; $i++)
        for ($j = 0; $j < $n; $j++)
            $sum += $arr[$i][$j];
  
    return $sum;
}
  
    // Driver Code
    $n = 3;
    echo findSum($n);
  
// This code is contributed by anuj_67.
?>


Output:

8

 
Method 2 (O(n)):
Consider n = 3, matrix formed will be:
0 1 2
1 0 1
2 1 0



Observe, the main diagonal is always 0 since all i are equal to j. The diagonal just above and just below will always be 1 because at each cell either i is 1 greater than j or j is 1 greater than i and so on.
Following the pattern we can see that the total sum of all the elements in the matrix will be, for each i from 0 to n, add i*(n-i)*2.
Below is the implementation of above idea :

C++

// C++ program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
#include<bits/stdc++.h>
using namespace std;
  
// Retuen the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i*(n-i);
    return 2*sum;
}
  
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
import java.io.*;
  
class GFG {
  
// Retuen the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
static int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i * (n - i);
    return 2 * sum;
}
  
    // Driver Code
    static public void main(String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
  
// This code is contributed by vt_m.

C#

// C# program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
using System;
  
class GFG {
  
// Retuen the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
static int findSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += i * (n - i);
    return 2 * sum;
}
  
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
  
// This code is contributed by vt_m.

Python3

# Python 3 program to find sum 
# of matrix in which each element 
# is absolute difference of its 
# corresponding row and column 
# number row. 
  
# Return the sum of matrix in 
# which each element is absolute 
# difference of its corresponding
# row and column number row 
def findSum(n):
    sum = 0
    for i in range(n):
        sum += i * (n - i)
    return 2 * sum
  
# Driver code
n = 3
print(findSum(n))
  
# This code is contributed by Shrikant13

PHP

<?php
// PHP program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
  
// Return the sum of matrix in which each
// element is absolute difference of its
// corresponding row and column number row
function findSum($n)
{
    $sum = 0;
    for ( $i = 0; $i < $n; $i++)
        $sum += $i * ($n - $i);
    return 2 * $sum;
}
  
    // Driver Code
    $n = 3;
    echo findSum($n);
  
// This code is contributed by anuj_67.
?>


Output:

8

 

Method 3 (Trick):
Consider n = 3, matrix formed will be:
0 1 2
1 0 1
2 1 0

So, sum = 1 + 1 + 1 + 1 + 2 + 2.
On Rearranging, 1 + 2 + 1 + 2 + 2 = 1 + 2 + 1 + 22.
So, in every case we can rearrange the sum of matrix so that the answer always will be sum of first n – 1 natural number and sum of square of first n – 1 natural number.

Sum of first n natural number = ((n)*(n + 1))/2.
Sum of first n natural number = ((n)*(n + 1)*(2*n + 1)/6.

Below is the implementation of above idea :

C++

// C++ program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
#include<bits/stdc++.h>
using namespace std;
  
// Retuen the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n*(n+1))/2;
    sum += (n*(n+1)*(2*n + 1))/6;
    return sum;
}
  
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
import java.io.*;
  
public class GFG {
      
// Retuen the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
static int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n * (n + 1)) / 2;
    sum += (n * (n + 1) * (2 * n + 1)) / 6;
    return sum;
}
  
    // Driver Code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(findSum(n));
    }
}
  
// This code is contributed by vt_m.

Python3

# Python 3 program to find sum of matrix 
# in which each element is absolute 
# difference of its corresponding row 
# and column number row. 
  
# Return the sum of matrix in which 
# each element is absolute difference 
# of its corresponding row and column 
# number row 
def findSum(n):
    n -= 1
    sum = 0
    sum += (n * (n + 1)) / 2
    sum += (n * (n + 1) * (2 * n + 1)) / 6
    return int(sum
  
# Driver Code
n = 3
print(findSum(n)) 
  
# This code contributed by Rajput-Ji

C#

// C# program to find sum of matrix in which
// each element is absolute difference of its
// corresponding row and column number row.
using System;
  
public class GFG {
      
// Retuen the sum of matrix in which each element
// is absolute difference of its corresponding
// row and column number row
static int findSum(int n)
{
    n--;
    int sum = 0;
    sum += (n * (n + 1)) / 2;
    sum += (n * (n + 1) * (2 * n + 1)) / 6;
    return sum;
}
  
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 3;
        Console.WriteLine(findSum(n));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum of 
// matrix in which each element 
// is absolute difference of its 
// corresponding row and column 
// number row.
  
// Retuen the sum of matrix in 
// which each element is absolute 
// difference of its corresponding
// row and column number row
function findSum($n)
{
    $n--;
    $sum = 0;
    $sum += ($n * ($n + 1)) / 2;
    $sum += ($n * ($n + 1) * 
                  (2 * $n + 1)) / 6;
    return $sum;
}
  
// Driver Code
$n = 3;
echo findSum($n) ;
  
// This code is contributed
// by nitin mittal. 
?>


Output :

8

Source:
https://stackoverflow.com/questions/42043708/sum-of-matrix-in-which-each-element-is-absolute-difference-of-row-and-column

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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