Sum of matrix element where each elements is integer division of row and column

Consider a N X N matrix where each element is divided by a column number (integer division), i.e. mat[i][j] = floor((i+1)/(j+1)) where 0 <= i < n and 0 <= j < n. The task is to find the sum of all matrix elements.

Examples :

Input  : N = 2
Output : 4
2 X 2 matrix with given constraint:
1 0
2 1
Sum of matrix element: 4

Input  : N = 3
Output : 9

Method 1 (Brute Force): Run two loops, one for the row and another for the column, and find the integer part of (i / j) and add to the answer.
Below is the implementation of this approach:  

C++

// C++ program to find sum of matrix element
// where each element is integer division of
// row and column.
#include<bits/stdc++.h>
using namespace std;

// Return sum of matrix element where each element
// is division of its corresponding row and column.
int findSum(int n)
{
    int ans = 0;
    for (int i = 1; i <= n; i++)   // for rows
        for (int j = 1; j <= n; j++) // for columns
            ans += (i/j);
    return ans;
}

// Driven Program
int main()
{
    int N = 2;
    cout << findSum(N) << endl;
    return 0;
}

Java

// java program to find sum of matrix
// element where each element is integer
// division of row and column.

import java.io.*;

class GFG {

    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    static int findSum(int n)
    {
        int ans = 0;

        // for rows
        for (int i = 1; i <= n; i++)

            // for columns
            for (int j = 1; j <= n; j++)
                ans += (i/j);

        return ans;
    }

    // Driven Program
    public static void main (String[] args)
    {
        int N = 2;
        System.out.println( findSum(N));
    }
}

// This code is contributed by anuj_67.

Python3

# Python 3 program to find sum of
# matrix element where each element
# is integer division of row and column.

# Return sum of matrix element
# where each element is division
# of its corresponding row and column.
def findSum(N):
    ans = 0
    for i in range(1, N + 1):
        for j in range(1, N + 1):
            ans += i // j
    return ans

# Driver code
N = 2
print(findSum(N))

# This code is contributed
# by Shrikant13

C#

// C# program to find the sum of matrix
// element where each element is an integer
// division of row and column.
using System;

class GFG {

    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    static int findSum(int n)
    {
        int ans = 0;

        // for rows
        for (int i = 1; i <= n; i++)

            // for columns
            for (int j = 1; j <= n; j++)
                ans += (i/j);

        return ans;
    }

    // Driven Program
    public static void Main ()
    {
        int N = 2;
        Console.WriteLine( findSum(N));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find sum of matrix element
// where each element is integer division of
// row and column.

// Return sum of matrix element
// where each element is division
// of its corresponding row and
// column.
function findSum( $n)
{
    $ans = 0;

    // for rows
    for($i = 1; $i <= $n; $i++)

        // for columns
        for($j = 1; $j <= $n; $j++)
            $ans += ($i / $j);
    return floor($ans);
}

    // Driver Code
    $N = 2;
    echo findSum($N);

// This code is contributed by anuj_67.
?>

Javascript

<script>
    // Javascript program to find the sum of matrix
    // element where each element is an integer
    // division of row and column.

    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    function findSum(n)
    {
        let ans = 0;

        // for rows
        for (let i = 1; i <= n; i++)

            // for columns
            for (let j = 1; j <= n; j++)
                ans += parseInt(i/j, 10);

        return ans;
    }

    let N = 2;
    document.write(findSum(N));

</script>

Output

4

Time complexity: O(n²).
Auxiliary Space: O(1)

Method 2 (Efficient):

Let N = 9, the matrix will be  

Observe, for each jth column 
mat[i][k] = 0, for 1 <= k < j, 1 <= i <= N 
mat[i][k] = 1, for j <= k < 2*j, 1 <= i <= N 
mat[i][k] = 2, for 2*j <= k < 3*j, 1 <= i <= N 
and so on. 
So, in each column i, there are i - 1 zero, 
followed by i times 1, followed by i times 2, and so on.
We traverse matrix column by column and sum elements.

Below is the implementation of this approach.  

C++

// C++ program to find sum of matrix element
// where each element is integer division of
// row and column.
#include<bits/stdc++.h>
using namespace std;

// Return sum of matrix element where each
// element is division of its corresponding
// row and column.
int findSum(int n)
{
    int ans = 0, temp = 0, num;

    // For each column.
    for (int i = 1; i <= n && temp < n; i++)
    {
        // count the number of elements of
        // each column. Initialize to i -1
        // because number of zeroes are i - 1.
        temp = i - 1;

        // For multiply
        num = 1;

        while (temp < n)
        {
            if (temp + i <= n)
                ans += (i * num);
            else
                ans += ((n - temp) * num);

            temp += i;
            num ++;
        }
    }

    return ans;
}

// Driven Program
int main()
{
    int N = 2;
    cout << findSum(N) << endl;
    return 0;
}

Java

// java program to find sum of matrix element
// where each element is integer division of
// row and column.

import java.io.*;

class GFG {

    // Return sum of matrix element where each
    // element is division of its corresponding
    // row and column.
    static int findSum(int n)
    {
        int ans = 0, temp = 0, num;

        // For each column.
        for (int i = 1; i <= n && temp < n; i++)
        {

            // count the number of elements of
            // each column. Initialize to i -1
            // because number of zeroes are i - 1.
            temp = i - 1;

            // For multiply
            num = 1;

            while (temp < n)
            {
                if (temp + i <= n)
                    ans += (i * num);
                else
                    ans += ((n - temp) * num);

                temp += i;
                num ++;
            }
        }

        return ans;
    }

    // Driven Program
    public static void main (String[] args)
    {
        int N = 2;
        System.out.println(findSum(N));
    }
}

// This code is contributed by anuj_67.

Python3

# Program to find sum of matrix element
# where each element is integer division
# of row and column.

# Return sum of matrix element where each
# element is division of its corresponding
# row and column.
def findSum(n):
    ans = 0; temp = 0;

    for i in range(1, n + 1):

        # count the number of elements of
        # each column. Initialize to i -1
        # because number of zeroes are i - 1.
        if temp < n:
            temp = i - 1

            # For multiply
            num = 1
            while temp < n:
                if temp + i <= n:
                    ans += i * num
                else:
                    ans += (n - temp) * num
                temp += i
                num += 1
    return ans

# Driver Code
N = 2
print(findSum(N))

# This code is contributed by Shrikant13

C#

// C# program to find sum of matrix
// element where each element is
// integer division of row and column.
using System;

class GFG
{

    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and column.
    static int findSum(int n)
    {
        int ans = 0, temp = 0, num;

        // For each column.
        for (int i = 1; i <= n && temp < n; i++)
        {

            // count the number of elements
            // of each column. Initialize
            // to i -1 because number of
            // zeroes are i - 1.
            temp = i - 1;

            // For multiply
            num = 1;

            while (temp < n)
            {
                if (temp + i <= n)
                    ans += (i * num);
                else
                    ans += ((n - temp) * num);

                temp += i;
                num ++;
            }
        }

        return ans;
    }

    // Driver Code
    public static void Main ()
    {
        int N = 2;
        Console.WriteLine(findSum(N));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find sum of
// matrix element where each
// element is integer division
// of row and column.

// Return sum of matrix element
// where each element is division
// of its corresponding row and column.
function findSum( $n)
{
    $ans = 0; $temp = 0; $num;

    // For each column.
    for ($i = 1; $i <= $n and
         $temp < $n; $i++)
    {
        // count the number of elements
        // of each column. Initialize
        // to i -1 because number of
        // zeroes are i - 1.
        $temp = $i - 1;

        // For multiply
        $num = 1;

        while ($temp < $n)
        {
            if ($temp + $i <= $n)
                $ans += ($i * $num);
            else
                $ans += (($n - $temp) *
                                $num);

            $temp += $i;
            $num ++;
        }
    }

    return $ans;
}

// Driver Code
$N = 2;
echo findSum($N);

// This code is contributed by anuj_67.
?>

Javascript

<script>
// java Script program to find sum of matrix element
// where each element is integer division of
// row and column.



// Return sum of matrix element where each
// element is division of its corresponding
// row and column.
function findSum(n) {
    let ans = 0, temp = 0, num;

    // For each column.
    for (let i = 1; i <= n && temp < n; i++)
    {

        // count the number of elements of
        // each column. Initialize to i -1
        // because number of zeroes are i - 1.
        temp = i - 1;

        // For multiply
        num = 1;

        while (temp < n)
        {
            if (temp + i <= n)
                ans += (i * num);
            else
                ans += ((n - temp) * num);

            temp += i;
            num ++;
        }
    }

    return ans;
}

// Driven Program

    let N = 2;
    document.write(findSum(N));



// This code is contributed by sravan kumar G
</script>

Output

3

Time complexity: O(n²)
Auxiliary Space: O(1), since no extra space has been taken.