Frecuencia de un entero en la array dada usando Divide and Conquer

Dada una array no ordenada arr[] y un entero K , la tarea es contar las ocurrencias de K en la array dada utilizando el método Divide and Conquer .

Ejemplos: 

Entrada: arr[] = {1, 1, 2, 2, 2, 2, 3}, K = 1 
Salida: 2

Entrada: arr[] = {1, 1, 2, 2, 2, 2, 3}, K = 4 
Salida: 0

Enfoque: La idea es dividir la array en dos partes de igual tamaño y contar el número de ocurrencias de K en cada mitad y luego sumarlas.

• Divida la array en dos partes hasta que solo quede un elemento en la array.
• Compruebe si un solo elemento en la array es K o no. Si es K , devuelve 1 , de lo contrario , 0 .
• Sume los valores devueltos para cada uno de los elementos para encontrar la ocurrencia de K en toda la array.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implrmrntation of the approach
 
#include <iostream>
using namespace std;
 
// Function to return the frequency of x
// in the subarray arr[low...high]
int count(int arr[], int low, int high, int x)
{
 
    // If the subarray is invalid or the
    // element is not found
    if ((low > high)
        || (low == high && arr[low] != x))
        return 0;
 
    // If there's only a single element
    // which is equal to x
    if (low == high && arr[low] == x)
        return 1;
 
    // Divide the array into two parts and
    // then find the count of occurrences
    // of x in both the parts
    return count(arr, low,
                 (low + high) / 2, x)
           + count(arr, 1 + (low + high) / 2,
                   high, x);
}
 
// Driver code
int main()
{
    int arr[] = { 30, 1, 42, 5, 56, 3, 56, 9 };
    int n = sizeof(arr) / sizeof(int);
    int x = 56;
 
    cout << count(arr, 0, n - 1, x);
 
    return 0;
}

// Java implrmrntation of the approach
 
class GFG {
 
    // Function to return the frequency of x
    // in the subarray arr[low...high]
    static int count(int arr[], int low,
                     int high, int x)
    {
 
        // If the subarray is invalid or the
        // element is not found
        if ((low > high)
            || (low == high && arr[low] != x))
            return 0;
 
        // If there's only a single element
        // which is equal to x
        if (low == high && arr[low] == x)
            return 1;
 
        // Divide the array into two parts and
        // then find the count of occurrences
        // of x in both the parts
        return count(arr, low,
                     (low + high) / 2, x)
            + count(arr, 1 + (low + high) / 2,
                    high, x);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 30, 1, 42, 5, 56, 3, 56, 9 };
        int n = arr.length;
        int x = 56;
        System.out.print(count(arr, 0, n - 1, x));
    }
}

# Python3 implrmrntation of the approach
 
# Function to return the frequency of x
# in the subarray arr[low...high]
def count(arr, low, high, x):
 
    # If the subarray is invalid or the
    # element is not found
    if ((low > high) or (low == high and arr[low] != x)):
        return 0;
 
    # If there's only a single element
    # which is equal to x
    if (low == high and arr[low] == x):
        return 1;
 
    # Divide the array into two parts and
    # then find the count of occurrences
    # of x in both the parts
    return count(arr, low, (low + high) // 2, x) +\
    count(arr, 1 + (low + high) // 2, high, x);
 
# Driver code
if __name__ == '__main__':
    arr = [ 30, 1, 42, 5, 56, 3, 56, 9];
    n = len(arr);
    x = 56;
    print(count(arr, 0, n - 1, x));
 
# This code is contributed by PrinciRaj1992

// C# implrmrntation of the approach
using System;
 
class GFG
{
 
    // Function to return the frequency of x
    // in the subarray arr[low...high]
    static int count(int []arr, int low,
                    int high, int x)
    {
 
        // If the subarray is invalid or the
        // element is not found
        if ((low > high)
            || (low == high && arr[low] != x))
            return 0;
 
        // If there's only a single element
        // which is equal to x
        if (low == high && arr[low] == x)
            return 1;
 
        // Divide the array into two parts and
        // then find the count of occurrences
        // of x in both the parts
        return count(arr, low,
                    (low + high) / 2, x)
            + count(arr, 1 + (low + high) / 2,
                    high, x);
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 30, 1, 42, 5, 56, 3, 56, 9 };
        int n = arr.Length;
        int x = 56;
        Console.Write(count(arr, 0, n - 1, x));
    }
}
 
// This code is contributed by AnkitRai01

<script>
// Javascript implrmrntation of the approach
// Function to return the frequency of x
// in the subarray arr[low...high]
 
function count(arr, low, high, x) {
 
    // If the subarray is invalid or the
    // element is not found
    if ((low > high)
        || (low == high && arr[low] != x))
        return 0;
 
    // If there's only a single element
    // which is equal to x
    if (low == high && arr[low] == x)
        return 1;
 
    // Divide the array into two parts and
    // then find the count of occurrences
    // of x in both the parts
    return count(arr, low,
        Math.floor((low + high) / 2), x)
        + count(arr, 1 + Math.floor((low + high) / 2),
            high, x);
}
 
// Driver code
 
let arr = [30, 1, 42, 5, 56, 3, 56, 9];
let n = arr.length;
let x = 56;
 
document.write(count(arr, 0, n - 1, x));
 
// This code is contributed by _saurabh_jaiswal
</script>

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Complejidad de tiempo: O (NlogN)