# Count array elements whose highest power of 2 less than or equal to that number is present in the given array

• Last Updated : 20 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to find the number of array elements whose highest power of 2 less than or equal to that number is present in the array.

Examples:

Input: arr[] = {3, 4, 6, 9}
Output: 2
Explanation:
There are 2 array elements (4 and 6), whose highest power of 2 is less than it (i.e. 4) are present in the array.

Input: arr[] = {3, 9, 10, 8, 1}
Output: 3

Naive Approach: The given problem can be solved by count those elements whose highest power of 2 exists in the given array and that can be found by traversing the array again. After checking for all the elements, print the total count obtained.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using an unordered_map to keep the count of visited elements and update the count of resultant elements accordingly. Follow the steps below to solve the given problem:

• Initialize a variable, say count that stores the count of elements whose highest power of 2 less than or equals to that is present in the array.
• Initialize a map M and store the frequency of array elements.
• Traverse the given array and for each element, if the frequency of the highest power of 2 of arr[i] not exceeding the element exists in the map then increment the value of count by 1.
• After completing the above steps, print the value of count as the resultant count of elements.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to count array elements``// whose highest power of 2 is less``// than or equal to that number is``// present in the given array``int` `countElement(``int` `arr[], ``int` `N)``{``    ``// Stores the resultant count``    ``// of array elements``    ``int` `count = 0;` `    ``// Stores frequency of``    ``// visited array elements``    ``unordered_map<``int``, ``int``> m;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {``        ``m[arr[i]]++;``    ``}` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Calculate log base 2``        ``// of the element arr[i]``        ``int` `lg = log2(arr[i]);` `        ``// Highest power of 2 whose``        ``// value is at most arr[i]``        ``int` `p = ``pow``(2, lg);` `        ``// Increment the count by 1``        ``if` `(m[p]) {``            ``count++;``        ``}``    ``}` `    ``// Return the resultant count``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 3, 4, 6, 9 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << countElement(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.HashMap;``import` `java.io.*;` `class` `GFG{``    ` `static` `int` `log2(``int` `N)``{``    ` `    ``// Calculate log2 N indirectly``    ``// using log() method``    ``int` `result = (``int``)(Math.log(N) /``                       ``Math.log(``2``));``    ` `    ``return` `result;``}` `// Function to count array elements``// whose highest power of 2 is less``// than or equal to that number is``// present in the given array``static` `int` `countElement(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores the resultant count``    ``// of array elements``    ``int` `count = ``0``;` `    ``// Stores frequency of``    ``// visited array elements``   ``HashMap m = ``new` `HashMap<>();``   ` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``if` `(m.containsKey(arr[i]))``        ``{``            ``m.put(arr[i], m.get(arr[i]) + ``1``);``        ``}``        ``else``        ``{``            ``m.put(arr[i], ``1``);``        ``}``    ``}` `    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// Calculate log base 2``        ``// of the element arr[i]``        ``int` `lg = log2(arr[i]);` `        ``// Highest power of 2 whose``        ``// value is at most arr[i]``        ``int` `p = (``int``)Math.pow(``2``, lg);` `        ``// Increment the count by 1``        ``if` `(m.containsKey(p))``        ``{``            ``count++;``        ``}``    ``}` `    ``// Return the resultant count``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = { ``3``, ``4``, ``6``, ``9` `};``    ``int` `N = arr.length;``    ` `    ``System.out.println(countElement(arr, N));``}``}` `// This code is contributed by Potta Lokesh`

## Python3

 `# Python program for the above approach``from` `math ``import` `log2` `# Function to count array elements``# whose highest power of 2 is less``# than or equal to that number is``# present in the given array``def` `countElement(arr, N):``    ``# Stores the resultant count``    ``# of array elements``    ``count ``=` `0` `    ``# Stores frequency of``    ``# visited array elements``    ``m ``=` `{}` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):``        ``m[arr[i]] ``=` `m.get(arr[i], ``0``) ``+` `1`  `    ``for` `i ``in` `range``(N):``        ``# Calculate log base 2``        ``# of the element arr[i]``        ``lg ``=` `int``(log2(arr[i]))` `        ``# Highest power of 2 whose``        ``# value is at most arr[i]``        ``p ``=` `pow``(``2``, lg)` `        ``# Increment the count by 1``        ``if` `(p ``in` `m):``            ``count ``+``=` `1` `    ``# Return the resultant count``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr``=` `[``3``, ``4``, ``6``, ``9``]``    ``N ``=` `len``(arr)``    ``print` `(countElement(arr, N))` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach` `using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to count array elements``// whose highest power of 2 is less``// than or equal to that number is``// present in the given array``static` `int` `countElement(``int` `[]arr, ``int` `N)``{``    ``// Stores the resultant count``    ``// of array elements``    ``int` `count = 1;` `    ``// Stores frequency of``    ``// visited array elements``    ``Dictionary<``int``,``int``> m = ``new` `Dictionary<``int``,``int``>();` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {``        ``if``(m.ContainsKey(arr[i]))``         ``m[arr[i]]++;``        ``else``         ``m.Add(arr[i],1);``    ``}` `    ``for``(``int` `i = 0; i < N; i++) {` `        ``// Calculate log base 2``        ``// of the element arr[i]``        ``int` `lg = (``int``)Math.Log(arr[i]);` `        ``// Highest power of 2 whose``        ``// value is at most arr[i]``        ``int` `p = (``int``)Math.Pow(2, lg);` `        ``// Increment the count by 1``        ``if` `(m.ContainsKey(p)) {``            ``count++;``        ``}``    ``}` `    ``// Return the resultant count``    ``return` `count;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 3, 4, 6, 9 };``    ``int` `N = arr.Length;``    ``Console.Write(countElement(arr, N));` `}``}` `// This code is contributed by bgangwar59.`

## Javascript

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Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(N)

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