# Highest powers of 2 not exceeding non-repeating array elements

• Difficulty Level : Easy
• Last Updated : 25 Nov, 2021

Given an array arr[] of size N, the task is for every non-repeating array element is to find the highest power of 2 that does not exceed that element. Print the powers of 2 in ascending order. If the array does not contain any non-repeating element, print “0”.

Examples:

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Input: arr[ ] = { 4, 5, 4, 3, 3, 4 }
Output:
Explanation: The only non-repeating element in the array is 5. Therefore, the highest power of 2 not exceeding 5 is 4.

Input: arr[ ] = { 1, 1, 7, 6, 3 }
Output: 2 4 4

Naive Approach: The simplest approach to solve this problem is to traverse the array and for each array element, check if it is non-repeating or not. For elements douns to be non-repeating, add them into another array. Then, for each element in the new array, find the highest powers of 2 not exceeding that element and print them in ascending order.
Time Complexity: O(N2 * log arr[i]), where arr[i] is the largest number of the array.
Auxiliary Space: O(N)

Efficient Approach: The optimal idea is to use Hashing. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to find the highest power of 2 for``// every non-repeating element in the array``void` `uniqueElement(``int` `arr[], ``int` `N)``{` `    ``// Stores the frequency``    ``// of array elements``    ``unordered_map<``int``, ``int``> freq;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Update frequency of each``        ``// element of the array``        ``freq[arr[i]]++;``    ``}` `    ``// Stores the non-repeating``    ``// array elements``    ``vector<``int``> v;` `    ``// Traverse the Map``    ``for` `(``auto` `i : freq) {` `        ``if` `(i.second == 1) {` `            ``// Calculate log base 2``            ``// of the current element``            ``int` `lg = log2(i.first);` `            ``// Highest power of 2 <= i.first``            ``int` `p = ``pow``(2, lg);` `            ``// Insert it into the vector``            ``v.push_back(p);``        ``}``    ``}` `    ``// If no element is non-repeating``    ``if` `(v.size() == 0) {``        ``cout << ``"0"``;``        ``return``;``    ``}` `    ``// Sort the powers of 2 obtained``    ``sort(v.begin(), v.end());` `    ``// Print the elements in the vector``    ``for` `(``auto` `i : v)``        ``cout << i << ``" "``;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 4, 5, 4, 3, 3, 4 };` `    ``// Size of array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``uniqueElement(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG{` `  ``// Function to find the highest power of 2 for``  ``// every non-repeating element in the array``  ``static` `void` `uniqueElement(``int` `arr[], ``int` `N)``  ``{` `    ``// Stores the frequency``    ``// of array elements``    ``HashMap freq``      ``= ``new` `HashMap();` `    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``      ``if` `(freq.containsKey(arr[i]))``      ``{``        ``freq.put(arr[i], freq.get(arr[i]) + ``1``);``      ``}``      ``else``      ``{``        ``freq.put(arr[i], ``1``);``      ``}``    ``}` `    ``// Stores the non-repeating``    ``// array elements``    ``ArrayList v``      ``= ``new` `ArrayList();` `    ``// Traverse the Map``    ``for` `(Map.Entry i : freq.entrySet()) {` `      ``if` `((``int``)i.getValue() == ``1``) {` `        ``// Calculate log base 2``        ``// of the current element``        ``int` `lg = (``int``)(Math.log((``int``)i.getKey()) / Math.log(``2``));` `        ``// Highest power of 2 <= i.first``        ``int` `p = (``int``)Math.pow(``2``, lg);` `        ``// Insert it into the vector``        ``v.add(p);``      ``}``    ``}` `    ``// If no element is non-repeating``    ``if` `(v.size() == ``0``) {``      ``System.out.print(``"0"``);``      ``return``;``    ``}` `    ``// Sort the powers of 2 obtained``    ``Collections.sort(v);` `    ``// Print the elements in the vector``    ``for` `(``int` `i : v)``      ``System.out.print( i + ``" "``);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``4``, ``5``, ``4``, ``3``, ``3``, ``4` `};` `    ``// Size of array``    ``int` `N = arr.length;``    ``uniqueElement(arr, N);``  ``}``}` `// This code is contributed by code_hunt.`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to find the highest power of 2 for``# every non-repeating element in the array``def` `uniqueElement(arr, N):` `    ``# Stores the frequency``    ``# of array elements``    ``freq ``=` `{}` `    ``# Traverse the array``    ``for` `i ``in` `range``(N) :`` ` `        ``# Update frequency``        ``# of arr[i]``        ``if` `arr[i] ``in` `freq :``            ``freq[arr[i]] ``+``=` `1``;``        ``else` `:``            ``freq[arr[i]] ``=` `1``;``    ` `    ``# Stores the non-repeating``    ``# array elements``    ``v ``=` `[]` `    ``# Traverse the Map``    ``for` `i ``in` `freq :``        ``if` `(freq[i] ``=``=` `1``) :` `            ``# Calculate log base 2``            ``# of the current element``            ``lg ``=` `int``(math.log2(i))` `            ``# Highest power of 2 <= i.first``            ``p ``=` `pow``(``2``, lg)` `            ``# Insert it into the vector``            ``v.append(p)``        ` `    ``# If no element is non-repeating``    ``if` `(``len``(v) ``=``=` `0``) :``        ``print``(``"0"``)``        ``return``    ` `    ``# Sort the powers of 2 obtained``    ``v.sort()` `    ``# Print elements in the vector``    ``for` `i ``in` `v :``        ``print``(i, end ``=` `" "``)` `# Driver Code``arr ``=` `[ ``4``, ``5``, ``4``, ``3``, ``3``, ``4` `]` `# Size of array``N ``=` `len``(arr)``uniqueElement(arr, N)` `# This code is contributed by sanjoy_62.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG{` `  ``// Function to find the highest power of 2 for``  ``// every non-repeating element in the array``  ``static` `void` `uniqueElement(``int` `[]arr, ``int` `N)``  ``{` `    ``// Stores the frequency``    ``// of array elements``    ``Dictionary<``int``, ``int``> freq``      ``= ``new` `Dictionary<``int``, ``int``>();` `    ``for` `(``int` `i = 0; i < N; i++)``    ``{``      ``if` `(freq.ContainsKey(arr[i]))``      ``{``        ``freq[arr[i]] = freq[arr[i]] + 1;``      ``}``      ``else``      ``{``        ``freq.Add(arr[i], 1);``      ``}``    ``}` `    ``// Stores the non-repeating``    ``// array elements``    ``List<``int``> v``      ``= ``new` `List<``int``>();` `    ``// Traverse the Map``    ``foreach``(KeyValuePair<``int``, ``int``> i ``in` `freq) {` `      ``if` `((``int``)i.Value == 1) {` `        ``// Calculate log base 2``        ``// of the current element``        ``int` `lg = (``int``)(Math.Log((``int``)i.Key) / Math.Log(2));` `        ``// Highest power of 2 <= i.first``        ``int` `p = (``int``)Math.Pow(2, lg);` `        ``// Insert it into the vector``        ``v.Add(p);``      ``}``    ``}` `    ``// If no element is non-repeating``    ``if` `(v.Count == 0) {``      ``Console.Write(``"0"``);``      ``return``;``    ``}` `    ``// Sort the powers of 2 obtained``    ``v.Sort();` `    ``// Print the elements in the vector``    ``foreach` `(``int` `i ``in` `v)``      ``Console.Write( i + ``" "``);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int` `[]arr = { 4, 5, 4, 3, 3, 4 };` `    ``// Size of array``    ``int` `N = arr.Length;``    ``uniqueElement(arr, N);``  ``}``}`  `// This code is contributed by 29AjayKumar`

## Javascript

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Output:
`4`

Time Complexity: O(N * log(MAXM)), where MAXM is the largest element present the array.
Auxiliary Space: O(N)

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