Rearrange array to make decimal equivalents of reversed binary representations of array elements sorted
Last Updated :
23 Jul, 2025
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:
- 43 --> (101011)2 --> reversed --> 53.
- 52 --> (110100)2 --> reversed --> 11.
- 61 --> (111101)2 --> reversed --> 47.
- 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 <bits/stdc++.h>
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<vector<int>> getNew(vector<int> arr)
{
// Stores the new array elements
vector<vector<int>> ans;
for (int i:arr)
ans.push_back({keyFunc(i), i});
return ans;
}
// Function for rearranging the array
vector<int> getArr(vector<vector<int> > 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<vector<int> > 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 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 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)
// C# program for the above approach
using System;
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][];
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;
foreach(var i in 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
Array.Sort(
newArr, new Comparison<int[]>((a, b) => {
return (a[0] == b[0]) ? (a[1] - b[1])
: (a[0] - b[0]);
}));
// Get arr from newArr
arr = getArr(newArr);
// Print the sorted array
int n = arr.Length;
Console.Write("[");
for (int i = 0; i < n - 1; i++)
Console.Write(arr[i] + ", ");
Console.Write(arr[n - 1] + "]");
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 43, 52, 61, 41 };
sortArray(arr);
}
}
// This code is contributed by phasing17
<script>
// JavaScript program for the above approach
// Function to reverse the bits of a number
function keyFunc(n) {
// Stores the reversed number
let 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
function getNew(arr) {
// Stores the new array elements
let ans = [];
for (let i of arr)
ans.push([keyFunc(i), i]);
return ans;
}
// Function for rearranging the array
function getArr(arr) {
// Stores the new array
let ans = [];
for (let i of arr)
ans.push(i[1]);
return ans;
}
// Function to sort the array by reversing
// binary representation
function sortArray(arr) {
// Creating a new array
let newArr = getNew(arr);
// Sort the array with the key
newArr.sort();
// Get arr from newArr
arr = getArr(newArr);
// Print the sorted array
let n = arr.length;
document.write("[");
for (let i = 0; i < n - 1; i++)
document.write(arr[i] + ", ");
document.write(arr[n - 1] + "]");
}
// Driver Code
let arr = [43, 52, 61, 41];
sortArray(arr);
// This code is contributed by _saurabh_jaiswal
</script>
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:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to reverse the bits of 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;
}
//defining the comparison function
bool compare(int a, int b)
{
return keyFunc(a) < keyFunc(b);
}
// Function to sort the array by reversing
// binary representation
void sortArray(int arr[], int n)
{
// Sort the array with the key
sort(arr, arr + n, compare);
// Print the sorted array
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
int main()
{
// Driver Code
int arr[] = { 43, 52, 61, 41 };
int n = 4;
sortArray(arr, n);
return 0;
}
// this code is contributed by phasing17
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
// Function to reverse the bits of number
public static int keyFunc(String num)
{
int n = Integer.parseInt(num);
// 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 to sort the array by reversing
// binary representation
public static void sortArray(List<Integer> arrInt,
int n)
{
// Sort the array with the key
List<String> arr = new ArrayList<String>();
for (int i = 0; i < arrInt.size(); i++) {
arr.add(Integer.toString(arrInt.get(i)));
}
Collections.sort(arr, new Comparator<String>() {
@Override
// defining the comparison function
public int compare(final String a, String b)
{
if (keyFunc(a) < keyFunc(b))
return -1;
return 1;
}
});
// Print the sorted array
for (int i = 0; i < n; i++)
System.out.print(arr.get(i) + " ");
}
// Driver Code
public static void main(String[] args)
{
List<Integer> arr = new ArrayList<Integer>();
arr.add(43);
arr.add(52);
arr.add(61);
arr.add(41);
int n = 4;
sortArray(arr, n);
}
}
// This code is contributed by akashish__
# 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)
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to reverse the bits of number
public 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;
}
// defining the comparison function
public static int compare(int a, int b)
{
return keyFunc(a) - keyFunc(b);
}
// Function to sort the array by reversing
// binary representation
public static void sortArray(int[] arr, int n)
{
// Sort the array with the key
Array.Sort(arr, compare);
// Print the sorted array
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
public static void Main(string[] args)
{
// Driver Code
int[] arr = { 43, 52, 61, 41 };
int n = 4;
sortArray(arr, n);
}
}
// This code is contributed by phasing17
// JavaScript program for the above approach
// Function to reverse the bits of number
function keyFunc(n)
{
// Stores the reversed number
var 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
function sortArray(arr)
{
// Sort the array with the key
arr.sort((a, b) => keyFunc(a) > keyFunc(b));
// Print the sorted array
console.log(arr);
}
// Driver Code
var arr = [43, 52, 61, 41];
sortArray(arr);
//this code is contributed by phasing17
Time Complexity: O(N * log N)
Auxiliary Space: O(1)
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