使用分而治之的给定数组中整数的频率

给定一个未排序的数组arr[]和一个整数K ，任务是使用分治法计算给定数组中K的出现次数。

例子：

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

做法：思路是将数组分成大小相等的两部分，计算每半部分K出现的次数，然后相加。

• 将数组分成两部分，直到数组中只剩下一个元素。
• 检查数组中的单个元素是否为K。如果是K则返回1否则返回0 。
• 将每个元素的返回值相加，以找出K在整个数组中的出现次数。

下面是上述方法的实现：

C++

// C++ implrmrntation of the approach
 
#include 
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

// 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

# 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#

// 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

Javascript

// 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

输出：

2

时间复杂度： O(NlogN)

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