Count array elements that divide the sum of all other elements
Given an array arr[], the task is to count the number of elements from the array which divide the sum of all other elements.
Examples:
Input: arr[] = {3, 10, 4, 6, 7}
Output: 3
3 divides (10 + 4 + 6 + 7) i.e. 27
10 divides (3 + 4 + 6 + 7) i.e. 20
6 divides (3 + 10 + 4 + 7) i.e. 24
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 2
Naive Approach: Run two-loop from 0 to N, calculate the sum of all elements except the current element and if this element divides that sum, then increment the count.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int countNum( int N, int arr[])
{
int count = 0;
for ( int i = 0; i < N; i++) {
int sum = 0;
for ( int j = 0; j < N; j++) {
if (i == j)
continue ;
else
sum += arr[j];
}
if (sum % arr[i] == 0)
count++;
}
return count;
}
int main()
{
int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countNum(n, arr);
return 0;
}
|
Java
class GFG
{
static int countNum( int N, int arr[])
{
int count = 0 ;
for ( int i = 0 ; i < N; i++)
{
int sum = 0 ;
for ( int j = 0 ; j < N; j++)
{
if (i == j)
continue ;
else
sum += arr[j];
}
if (sum % arr[i] == 0 )
count++;
}
return count;
}
public static void main(String[] args)
{
int arr[] = { 3 , 10 , 4 , 6 , 7 };
int n = arr.length;
System.out.println(countNum(n, arr));
}
}
|
Python3
def countNum(N, arr):
count = 0
for i in range (N):
Sum = 0
for j in range (N):
if (i = = j):
continue
else :
Sum + = arr[j]
if ( Sum % arr[i] = = 0 ):
count + = 1
return count
arr = [ 3 , 10 , 4 , 6 , 7 ]
n = len (arr)
print (countNum(n, arr))
|
C#
using System;
class GFG
{
static int countNum( int N, int []arr)
{
int count = 0;
for ( int i = 0; i < N; i++)
{
int sum = 0;
for ( int j = 0; j < N; j++)
{
if (i == j)
continue ;
else
sum += arr[j];
}
if (sum % arr[i] == 0)
count++;
}
return count;
}
public static void Main()
{
int []arr = { 3, 10, 4, 6, 7 };
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
}
}
|
PHP
<?php
function countNum( $N , $arr )
{
$count = 0;
for ( $i =0; $i < $N ; $i ++) {
$Sum = 0;
for ( $j = 0; $j < $N ; $j ++) {
if ( $i == $j )
continue ;
else
$Sum += $arr [ $j ];
}
if ( $Sum % $arr [ $i ] == 0)
$count += 1;
}
return $count ;
}
$arr = array (3, 10, 4, 6, 7);
$n = count ( $arr );
echo countNum( $n , $arr );
?>
|
Javascript
<script>
function countNum(N, arr)
{
let count = 0;
for (let i = 0; i < N; i++) {
let sum = 0;
for (let j = 0; j < N; j++) {
if (i == j)
continue ;
else
sum += arr[j];
}
if (sum % arr[i] == 0)
count++;
}
return count;
}
let arr = [ 3, 10, 4, 6, 7 ];
let n = arr.length;
document.write(countNum(n, arr));
</script>
|
Time Complexity: O(N2)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach: Run a single loop from 0 to N, calculate the sum of all the elements. Now run another loop from 0 to N and if (sum – arr[i]) % arr[i] = 0 then increment the count.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int countNum( int N, int arr[])
{
int sum = 0, count = 0;
for ( int i = 0; i < N; i++)
sum += arr[i];
for ( int i = 0; i < N; i++)
if ((sum - arr[i]) % arr[i] == 0)
count++;
return count;
}
int main()
{
int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countNum(n, arr);
return 0;
}
|
Java
class GFG
{
static int countNum( int N, int arr[])
{
int sum = 0 , count = 0 ;
for ( int i = 0 ; i < N; i++)
{
sum += arr[i];
}
for ( int i = 0 ; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0 )
{
count++;
}
}
return count;
}
public static void main(String[] args)
{
int arr[] = { 3 , 10 , 4 , 6 , 7 };
int n = arr.length;
System.out.println(countNum(n, arr));
}
}
|
Python3
def countNum(N, arr):
Sum , count = 0 , 0
for i in range (N):
Sum + = arr[i]
for i in range (N):
if (( Sum - arr[i]) % arr[i] = = 0 ):
count + = 1
return count
arr = [ 3 , 10 , 4 , 6 , 7 ]
n = len (arr)
print (countNum(n, arr))
|
C#
using System;
class GFG
{
static int countNum( int N, int []arr)
{
int sum = 0, count = 0;
for ( int i = 0; i < N; i++)
{
sum += arr[i];
}
for ( int i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
return count;
}
public static void Main()
{
int []arr = {3, 10, 4, 6, 7};
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
}
}
|
PHP
<?php
function countNum( $N , $arr )
{
$sum = 0; $count = 0;
for ( $i = 0; $i < $N ; $i ++)
$sum += $arr [ $i ];
for ( $i = 0; $i < $N ; $i ++)
if (( $sum - $arr [ $i ]) % $arr [ $i ] == 0)
$count ++;
return $count ;
}
$arr = array (3, 10, 4, 6, 7);
$n = count ( $arr );
echo countNum( $n , $arr );
?>
|
Javascript
<script>
function countNum(N, arr)
{
let sum = 0, count = 0;
for (let i = 0; i < N; i++)
{
sum += arr[i];
}
for (let i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
return count;
}
let arr = [3, 10, 4, 6, 7];
let n = arr.length;
document.write(countNum(n, arr));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
15 Nov, 2022
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