Count occurrences of the average of array elements with a given number

Given an array of integers and an integer . For every integer of the array a[i], the task is to calculate the count of numbers in the array with value equals to the average of element a[i] and x. That is, the number of occurrences of the (average of element a[i] and x) in the array.

Examples:

Input: arr[] = {2, 0, 4, 6, 2}, x = 2 
Output: 2 0 0 1 2 
For x = 2, the average values for 2, 0, 4, 6, 2 would be 2, 1, 3, 4, 2 respectively. So, the count array would result in 2, 0, 0, 1, 2. 

Input: arr[] = {9, 5, 2, 4, 0, 3}, x = 3 
Output: 0 1 1 1 0 1 
For x = 3, the average values for 9, 5, 2, 4, 0, 3 would be 6, 4, 2, 3, 1, 2 respectively. So, the count array would result in 0, 1, 1, 1, 0, 1. 

Approach: 

1. Traverse the array and map every element with its count of occurrence in the array.
2. Now traverse the array again, take the average of the array element and given and check for its value in the map.

Below is the implementation of the above approach:  

C++

// CPP program to find the count of
// occurrences of the average of array
// elements with a given number
#include<bits/stdc++.h>

using namespace std;


    // Function to find the count of
    // occurrences of the average of array
    // elements with a given number
    void getAverageCountArray(int a[], int x, int N)
    {
        // mp to store count of occurrence
        // of every array element in the array
        map<int,int> mp;

        // Array that stores the average
        // count for given array
        int avg[N] = {0};
        int val, av;

        for (int i = 0; i < N; i++)
        {
            // first occurrence of a[i]
            if (mp[a[i]] == 0)
                mp[a[i]] = 1;

            else
            mp[a[i]]++;


            // element has already occurred before
            // so increase its count
        }

        for (int i = 0; i < N; i++)
        {
            av = int((a[i] + x) / 2);
            if (mp.find(av) != mp.end())
            {
                val = mp[av];
                avg[i] = val;
            }
        }

        // Printing the average count array
        for (int i = 0; i < N; i++)
        {
            cout << avg[i] << " ";
        }
    }

    // Driver code
    int main()
    {
        int a[] = { 2, 0, 4, 6, 2 };
        int x = 2;

        int N = sizeof(a)/sizeof(a[0]);
        getAverageCountArray(a, x, N);
    }

// This code is contributed by
// Surendra_Gangwar

Java

// Java program to find the count of
// occurrences of the average of array
// elements with a given number
import java.io.*;
import java.util.*;
import java.lang.*;

class GFG {

    // Function to find the count of
    // occurrences of the average of array
    // elements with a given number
    static void getAverageCountArray(int[] a, int x, int N)
    {
        // Map to store count of occurrence
        // of every array element in the array
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

        // Array that stores the average
        // count for given array
        int[] avg = new int[N];
        int val, av;

        for (int i = 0; i < N; i++) {
            // first occurrence of a[i]
            if (!map.containsKey(a[i])) {
                map.put(a[i], 1);
            }

            // element has already occurred before
            // so increase its count
            else {
                // gives current count of a[i]
                val = map.get(a[i]);
                val++;
                map.remove(a[i]);
                map.put(a[i], val);
            }
        }

        for (int i = 0; i < N; i++) {
            av = (a[i] + x) / 2;
            if (map.containsKey(av)) {
                val = map.get(av);
                avg[i] = val;
            }
        }

        // Printing the average count array
        for (int i = 0; i < N; i++) {
            System.out.print(avg[i] + " ");
        }
    }

    // Driver code
    public static void main(String args[])
    {
        int[] a = { 2, 0, 4, 6, 2 };
        int x = 2;

        int N = a.length;

        getAverageCountArray(a, x, N);
    }
}

Python3

# Python3 program to find the count of
# occurrences of the average of array
# elements with a given number

# Function to find the count of
# occurrences of the average of array
# elements with a given number
def getAverageCountArray(a, x, N):

    # Dictionary to store count of occurrence
    # of every array element in the array
    map = {}

    # Array that stores the average
    # count for given array
    avg = [0] * N

    for i in range(N):
        # first occurrence of a[i]
        if a[i] not in map:
            map[a[i]] =  1

        # element has already occurred before
        # so increase its count
        else:
            # gives current count of a[i]
            map[a[i]] += 1

    for i in range(N):
        av = (a[i] + x) // 2

        if av in map:
            val = map[av]
            avg[i] = val

    # Printing the average count array
    for i in range(N):
        print(avg[i], end = " ")


if __name__ == "__main__":

    a = [2, 0, 4, 6, 2]
    x = 2

    N = len(a)
    getAverageCountArray(a, x, N)

# This code is contributed by Rituraj Jain

C#

// C# program to find the count of
// occurrences of the average of array
// elements with a given number
using System;
using System.Collections.Generic;

class GFG
{

// Function to find the count of
// occurrences of the average of array
// elements with a given number
static void getAverageCountArray(int[] a, int x,
                                          int N)
{
    // Map to store count of occurrence
    // of every array element in the array
    Dictionary<int,
               int> map = new Dictionary<int,
                                         int>();

    // Array that stores the average
    // count for given array
    int[] avg = new int[N];
    int val, av;

    for (int i = 0; i < N; i++)
    {
        // first occurrence of a[i]
        if (!map.ContainsKey(a[i]))
        {
            map.Add(a[i], 1);
        }

        // element has already occurred before
        // so increase its count
        else
        {
            // gives current count of a[i]
            val = map[a[i]];
            val++;
            map.Remove(a[i]);
            map.Add(a[i], val);
        }
    }

    for (int i = 0; i < N; i++)
    {
        av = (a[i] + x) / 2;
        if (map.ContainsKey(av))
        {
            val = map[av];
            avg[i] = val;
        }
    }

    // Printing the average count array
    for (int i = 0; i < N; i++)
    {
        Console.Write(avg[i] + " ");
    }
}

// Driver code
public static void Main()
{
    int[] a = { 2, 0, 4, 6, 2 };
    int x = 2;

    int N = a.Length;

    getAverageCountArray(a, x, N);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to find the count of
// occurrences of the average of array
// elements with a given number

    // Function to find the count of
    // occurrences of the average of array
    // elements with a given number
    function getAverageCountArray(a, x, N)
    {
        // Map to store count of occurrence
        // of every array element in the array
        let map = new Map();

        // Array that stores the average
        // count for given array
        let avg = Array.from({length: N}, (_, i) => 0);
        let val, av;

        for (let i = 0; i < N; i++) {
            // first occurrence of a[i]
            if (!map.has(a[i])) {
                map.set(a[i], 1);
            }

            // element has already occurred before
            // so increase its count
            else {
                // gives current count of a[i]
                val = map.get(a[i]);
                val++;
                map.delete(a[i]);
                map.set(a[i], val);
            }
        }

        for (let i = 0; i < N; i++) {
            av = (a[i] + x) / 2;
            if (map.has(av)) {
                val = map.get(av);
                avg[i] = val;
            }
        }

        // Printing the average count array
        for (let i = 0; i < N; i++) {
            document.write(avg[i] + " ");
        }
    }

// Driver Code

           let a = [ 2, 0, 4, 6, 2 ];
        let x = 2;

        let N = a.length;

        getAverageCountArray(a, x, N);

</script>

Output

2 0 0 1 2 

Time Complexity: O(n log n), where n is the size of the given array.
Auxiliary Space: O(n)