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. 24Input: 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++ implementation of the approach #include <iostream> using namespace std;
// Function to return the count // of the required numbers int countNum( int N, int arr[])
{ // To store the count of required numbers
int count = 0;
for ( int i = 0; i < N; i++) {
// Initialize sum to 0
int sum = 0;
for ( int j = 0; j < N; j++) {
// If current element and the
// chosen element are same
if (i == j)
continue ;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
} // Driver code int main()
{ int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countNum(n, arr);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to return the count // of the required numbers static int countNum( int N, int arr[])
{ // To store the count of required numbers
int count = 0 ;
for ( int i = 0 ; i < N; i++)
{
// Initialize sum to 0
int sum = 0 ;
for ( int j = 0 ; j < N; j++)
{
// If current element and the
// chosen element are same
if (i == j)
continue ;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0 )
count++;
}
// Return the count
return count;
} // Driver code public static void main(String[] args)
{ int arr[] = { 3 , 10 , 4 , 6 , 7 };
int n = arr.length;
System.out.println(countNum(n, arr));
} } // This code is contributed by Code_Mech |
# Python3 implementation of the approach # Function to return the count # of the required numbers def countNum(N, arr):
# To store the count of
# required numbers
count = 0
for i in range (N):
# Initialize sum to 0
Sum = 0
for j in range (N):
# If current element and the
# chosen element are same
if (i = = j):
continue
# Add all other numbers of array
else :
Sum + = arr[j]
# If Sum is divisible by the
# chosen element
if ( Sum % arr[i] = = 0 ):
count + = 1
# Return the count
return count
# Driver code arr = [ 3 , 10 , 4 , 6 , 7 ]
n = len (arr)
print (countNum(n, arr))
# This code is contributed # by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the count // of the required numbers static int countNum( int N, int []arr)
{ // To store the count of required numbers
int count = 0;
for ( int i = 0; i < N; i++)
{
// Initialize sum to 0
int sum = 0;
for ( int j = 0; j < N; j++)
{
// If current element and the
// chosen element are same
if (i == j)
continue ;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
} // Driver code public static void Main()
{ int []arr = { 3, 10, 4, 6, 7 };
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
} } // This code is contributed by inder_verma.. |
<?php // Php implementation of the approach // Function to return the count // of the required numbers function countNum( $N , $arr )
{ // To store the count of // required numbers $count = 0;
for ( $i =0; $i < $N ; $i ++) {
// Initialize sum to 0 $Sum = 0;
for ( $j = 0; $j < $N ; $j ++) {
// If current element and the
// chosen element are same
if ( $i == $j )
continue ;
// Add all other numbers of array
else
$Sum += $arr [ $j ];
}
// If Sum is divisible by the
// chosen element
if ( $Sum % $arr [ $i ] == 0)
$count += 1;
} // Return the count return $count ;
} // Driver code $arr = array (3, 10, 4, 6, 7);
$n = count ( $arr );
echo countNum( $n , $arr );
// This code is contributed // by Srathore ?> |
<script> // Javascript implementation of the approach
// Function to return the count
// of the required numbers
function countNum(N, arr)
{
// To store the count of required numbers
let count = 0;
for (let i = 0; i < N; i++) {
// Initialize sum to 0
let sum = 0;
for (let j = 0; j < N; j++) {
// If current element and the
// chosen element are same
if (i == j)
continue ;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
}
let arr = [ 3, 10, 4, 6, 7 ];
let n = arr.length;
document.write(countNum(n, arr));
// This code is contributed by vaibhavrabadiya117
</script> |
3
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++ implementation of the approach #include <iostream> using namespace std;
// Function to return the count // of the required numbers int countNum( int N, int arr[])
{ // Initialize sum and count to 0
int sum = 0, count = 0;
// Calculate sum of all
// the array elements
for ( int i = 0; i < N; i++)
sum += arr[i];
for ( int i = 0; i < N; i++)
// If current element satisfies the condition
if ((sum - arr[i]) % arr[i] == 0)
count++;
// Return the count of required elements
return count;
} // Driver code int main()
{ int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countNum(n, arr);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to return the count
// of the required numbers
static int countNum( int N, int arr[])
{
// Initialize sum and count to 0
int sum = 0 , count = 0 ;
// Calculate sum of all
// the array elements
for ( int i = 0 ; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for ( int i = 0 ; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0 )
{
count++;
}
}
// Return the count of required elements
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3 , 10 , 4 , 6 , 7 };
int n = arr.length;
System.out.println(countNum(n, arr));
}
} // This code has been contributed by 29AjayKumar |
# Python3 implementation of the approach # Function to return the count # of the required numbers def countNum(N, arr):
# Initialize Sum and count to 0
Sum , count = 0 , 0
# Calculate Sum of all the
# array elements
for i in range (N):
Sum + = arr[i]
for i in range (N):
# If current element satisfies
# the condition
if (( Sum - arr[i]) % arr[i] = = 0 ):
count + = 1
# Return the count of required
# elements
return count
# Driver code arr = [ 3 , 10 , 4 , 6 , 7 ]
n = len (arr)
print (countNum(n, arr))
# This code is contributed # by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the count
// of the required numbers
static int countNum( int N, int []arr)
{
// Initialize sum and count to 0
int sum = 0, count = 0;
// Calculate sum of all
// the array elements
for ( int i = 0; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for ( int i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
// Return the count of required elements
return count;
}
// Driver code
public static void Main()
{
int []arr = {3, 10, 4, 6, 7};
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
}
} /* This code contributed by PrinciRaj1992 */ |
<?php // PHP implementation of the approach // Function to return the count // of the required numbers function countNum( $N , $arr )
{ // Initialize sum and count to 0
$sum = 0; $count = 0;
// Calculate sum of all
// the array elements
for ( $i = 0; $i < $N ; $i ++)
$sum += $arr [ $i ];
for ( $i = 0; $i < $N ; $i ++)
// If current element satisfies
// the condition
if (( $sum - $arr [ $i ]) % $arr [ $i ] == 0)
$count ++;
// Return the count of required elements
return $count ;
} // Driver code $arr = array (3, 10, 4, 6, 7);
$n = count ( $arr );
echo countNum( $n , $arr );
// This code contributed by Rajput-Ji ?> |
<script> // Javascript implementation of the approach
// Function to return the count
// of the required numbers
function countNum(N, arr)
{
// Initialize sum and count to 0
let sum = 0, count = 0;
// Calculate sum of all
// the array elements
for (let i = 0; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for (let i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
// Return the count of required elements
return count;
}
let arr = [3, 10, 4, 6, 7];
let n = arr.length;
document.write(countNum(n, arr));
</script> |
3
Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.