Frequency of an integer in the given array using Divide and Conquer

Given an unsorted array arr[] and an integer K, the task is to count the occurrences of K in the given array using Divide and Conquer method.

Examples:

Input: arr[] = {1, 1, 2, 2, 2, 2, 3}, K = 1
Output: 2

Input: arr[] = {1, 1, 2, 2, 2, 2, 3}, K = 4
Output: 0

Approach: The idea is to divide the array into two parts of equal sizes and count the number of occurrences of K in each half and then add them up.



  • Divide the array into two parts until there is only one element left in the array.
  • Check that single element in the array is K or not, If it is K then return 1 otherwise 0.
  • Add up the returned values for each of the element to find the occurence of K in the whole array.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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Output:

2

Time Complexity: O(NlogN)

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