Sum of matrix element where each elements is integer division of row and column
Last Updated :
12 Sep, 2023
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 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.
# 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# 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 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.
?>
<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>
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 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.
# 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# 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 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.
?>
<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>
Time complexity: O(n2)
Auxiliary Space: O(1), since no extra space has been taken.
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