Dada una array arr[] , la tarea es contar el número de elementos de la array que dividen la suma de todos los demás elementos.
Ejemplos:
Entrada: arr[] = {3, 10, 4, 6, 7}
Salida: 3
3 divide (10 + 4 + 6 + 7) es decir, 27
10 divide (3 + 4 + 6 + 7) es decir, 20
6 divide (3 + 10 + 4 + 7) es decir, 24Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Salida: 2
Enfoque ingenuo: ejecute dos bucles de 0 a N, calcule la suma de todos los elementos excepto el elemento actual y, si este elemento divide esa suma, incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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#
// 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
// 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
?>
Javascript
<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>
Producción:
3
Complejidad temporal: O(N 2 )
Enfoque eficiente: Ejecute un solo ciclo de 0 a N, calcule la suma de todos los elementos. Ahora ejecuta otro ciclo de 0 a N y si (sum – arr[i]) % arr[i] = 0 entonces incrementa el conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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#
// 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
// 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
?>
Javascript
<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>
Producción:
3
Complejidad de tiempo: O(N)
Publicación traducida automáticamente
Artículo escrito por Vivek.Pandit y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA