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# Rearrange array to make decimal equivalents of reversed binary representations of array elements sorted

• Last Updated : 30 Jun, 2021

Given an array arr[] consisting of N positive integers, the task is to rearrange the array such that the reversed binary representation of all the array elements is sorted.

If the decimal equivalent of reversed binary representations of two or more array elements is equal, then the original value is taken into consideration while rearranging the array.

Examples:

Input: arr[] = {43, 52, 61, 41}
Output: 52 41 61 43
Explanation:
Below are the reversed binary representation of the array elements:

1. 43 –> (101011)2 –> reversed –> 53.
2. 52 –> (110100)2 –> reversed –> 11.
3. 61 –> (111101)2 –> reversed –> 47.
4. 41 –> (101001)2 –> reversed –> 37.

Therefore, after rearranging the array element as {52, 41, 61, 43}, the reversed binary representation of rearranged array elements is in sorted order.

Input: arr[] = {5, 3, 6, 2, 4}
Output: 2 4 3 6 5

Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to reverse the bits of a number``int` `keyFunc(``int` `n)``{` `    ``// Stores the reversed number``    ``int` `rev = 0;` `    ``while` `(n > 0)``    ``{` `        ``// Divide rev by 2``        ``rev = rev << 1;` `        ``// If the value of N is odd``        ``if` `(n & 1 == 1)``            ``rev = rev ^ 1;` `        ``// Update the value of N``        ``n = n >> 1;``      ``}` `    ``// Return the final value of rev``    ``return` `rev;``}` `// Function for rearranging the array``// element according to the given rules``vector> getNew(vector<``int``> arr)``{` `    ``// Stores the new array elements``    ``vector> ans;` `    ``for` `(``int` `i:arr)``        ``ans.push_back({keyFunc(i), i});` `    ``return` `ans;``}` `// Function for rearranging the array``vector<``int``> getArr(vector > arr){` `    ``// Stores the new array``    ``vector<``int``> ans;` `    ``for` `(``auto` `i:arr)``        ``ans.push_back(i[1]);``    ``return` `ans;``}`  `// Function to sort the array by reversing``// binary representation``void` `sortArray(vector<``int``> arr)``{` `  ``// Creating a new array``  ``vector > newArr = getNew(arr);` `  ``// Sort the array with the key``  ``sort(newArr.begin(),newArr.end());` `  ``// Get arr from newArr``  ``arr = getArr(newArr);` `  ``// Print the sorted array``  ``int` `n = arr.size();``  ``cout<<``"["``;``  ``for``(``int` `i = 0; i < n - 1; i++)``    ``cout << arr[i] << ``", "``;``  ``cout << arr[n - 1] << ``"]"``;``}` `// Driver Code``int` `main()``{``  ` `  ``vector<``int``> arr = {43, 52, 61, 41};``  ``sortArray(arr);` `  ``return` `0;``}` `// This code is contributed by mohit kumar 29.`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{` `// Function to reverse the bits of a number``static` `int` `keyFunc(``int` `n)``{``    ` `    ``// Stores the reversed number``    ``int` `rev = ``0``;` `    ``while` `(n > ``0``)``    ``{``        ` `        ``// Divide rev by 2``        ``rev = rev << ``1``;` `        ``// If the value of N is odd``        ``if` `((n & ``1``) == ``1``)``            ``rev = rev ^ ``1``;` `        ``// Update the value of N``        ``n = n >> ``1``;``    ``}` `    ``// Return the final value of rev``    ``return` `rev;``}` `// Function for rearranging the array``// element according to the given rules``static` `int``[][] getNew(``int` `arr[])``{``    ` `    ``// Stores the new array elements``    ``int` `ans[][] = ``new` `int``[arr.length][``2``];` `    ``for``(``int` `i = ``0``; i < arr.length; i++)``        ``ans[i] = ``new` `int``[]{ keyFunc(arr[i]), arr[i] };` `    ``return` `ans;``}` `// Function for rearranging the array``static` `int``[] getArr(``int``[][] arr)``{` `    ``// Stores the new array``    ``int` `ans[] = ``new` `int``[arr.length];``    ``int` `idx = ``0``;``    ``for``(``int` `i[] : arr)``        ``ans[idx++] = i[``1``];``        ` `    ``return` `ans;``}` `// Function to sort the array by reversing``// binary representation``static` `void` `sortArray(``int` `arr[])``{` `    ``// Creating a new array``    ``int``[][] newArr = getNew(arr);` `    ``// Sort the array with the key``    ``Arrays.sort(newArr, (a, b) -> {``        ``if` `(Integer.compare(a[``0``], b[``0``]) == ``0``)``            ``return` `Integer.compare(a[``1``], b[``1``]);``            ` `        ``return` `Integer.compare(a[``0``], b[``0``]);``    ``});` `    ``// Get arr from newArr``    ``arr = getArr(newArr);` `    ``// Print the sorted array``    ``int` `n = arr.length;``    ``System.out.print(``"["``);``    ``for``(``int` `i = ``0``; i < n - ``1``; i++)``        ``System.out.print(arr[i] + ``", "``);``        ` `    ``System.out.print(arr[n - ``1``] + ``"]"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``43``, ``52``, ``61``, ``41` `};``    ``sortArray(arr);``}``}` `// This code is contributed by Kingash`

## Python3

 `# Python3 program for the above approach` `# Function to reverse the bits of a number``def` `keyFunc(n):` `    ``# Stores the reversed number``    ``rev ``=` `0` `    ``while` `(n > ``0``):` `        ``# Divide rev by 2``        ``rev ``=` `rev << ``1` `        ``# If the value of N is odd``        ``if` `(n & ``1` `=``=` `1``):``            ``rev ``=` `rev ^ ``1` `        ``# Update the value of N``        ``n ``=` `n >> ``1` `    ``# Return the final value of rev``    ``return` `rev` `# Function for rearranging the array``# element according to the given rules``def` `getNew(arr):` `    ``# Stores the new array elements``    ``ans ``=` `[]` `    ``for` `i ``in` `arr:``        ``ans.append([keyFunc(i), i])``    ``return` `ans` `# Function for rearranging the array``def` `getArr(arr):` `    ``# Stores the new array``    ``ans ``=` `[]` `    ``for` `i ``in` `arr:``        ``ans.append(i[``1``])``    ``return` `ans` `# Function to sort the array by reversing``# the binary representation``def` `sortArray(arr):` `    ``# Creating a new array``    ``newArr ``=` `getNew(arr)` `    ``# Sort the array with the key``    ``newArr.sort()` `    ``# Get arr from newArr``    ``arr ``=` `getArr(newArr)` `    ``# Print the sorted array``    ``print``(arr)`  `# Driver Code` `arr ``=` `[``43``, ``52``, ``61``, ``41``]``sortArray(arr)`

## Javascript

 ``
Output:
`[52, 41, 61, 43]`

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

Space-Optimized Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## Python3

 `# Python3 program for the above approach` `# Function to reverse the bits of number``def` `keyFunc(n):``  ` `    ``# Stores the reversed number``    ``rev ``=` `0``    ` `    ``while` `(n > ``0``):``      ` `        ``# Divide rev by 2``        ``rev ``=` `rev << ``1``        ` `        ``# If the value of N is odd``        ``if` `(n & ``1` `=``=` `1``):``            ``rev ``=` `rev ^ ``1``            ` `        ``# Update the value of N``        ``n ``=` `n >> ``1``        ` `    ``# Return the final value of rev``    ``return` `rev` `# Function to sort the array by reversing``# binary representation``def` `sortArray(arr):` `    ``# Sort the array with the key``    ``arr ``=` `sorted``(arr, key ``=` `keyFunc)` `    ``# Print the sorted array``    ``print``(arr)` `# Driver Code` `arr ``=` `[``43``, ``52``, ``61``, ``41``]``sortArray(arr)`
Output:
`[52, 41, 61, 43]`

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

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