Reverse bits using lookup table in O(1) time

Given an unsigned integer, reverse all bits of it and return the number with reversed bits. Examples:

Input : n = 1
Output : 2147483648  
On a machine with size of unsigned
bit as 32. Reverse of 0....001 is
100....0.

Input : n = 2147483648
Output : 1       

In the previous post we had seen two method that solved this problem in O(n) & O(logn ) time. Here we solve this problem in O(1) time using lookup table. It's hard to reverse all 32 bits (assuming this as size of int) in one go using lookup table (" because it's infeasible to create lookup table of size 2³²-1 "). So we break 32 bits into 8 bits of chunks( lookup table of size 2⁸-1 "0-255"). Lookup Table in lookup tale we will store reverse of every number that are in a range( 0-255) LookupTable[0] = 0 | binary 00000000 Reverse 00000000 LookupTable[1] = 128 | binary 00000001 reverse 10000000 LookupTable[2] = 64 | binary 00000010 reverse 01000000 LookupTable[3] = 192 | binary 00000011 reverse 11000000 LookupTable[4] = 32 | binary 00000100 reverse 00100000 and so on... upto lookuptable[255]. Let's take an Example How lookup table work. let number = 12456 in Binary = 00000000000000000011000010101000

Split it into 8 bits chunks  :  00000000 | 00000000 | 00110000 | 10101000
         in decimal          :     0          0          48       168
reverse each chunks using lookup table :
Lookuptable[ 0 ] = 0  | in binary 00000000
Lookuptable[48 ] = 12 | in binary 00001100
Lookuptable[168] = 21 | in binary 00010101
 
Now Binary :  
00000000 | 00000000 | 00001100 | 00010101

Binary chunks after rearrangement : 
00010101 | 00001100 | 00000000 | 00000000   
  
Reverse of 12456 is 353107968  

// CPP program to reverse bits using lookup table.

#include <bits/stdc++.h>
using namespace std;

// Generate a lookup table for 32bit operating system
// using macro

#define R2(n) n, n + 2 * 64, n + 1 * 64, n + 3 * 64

#define R4(n)                                              \
    R2(n), R2(n + 2 * 16), R2(n + 1 * 16), R2(n + 3 * 16)

#define R6(n)                                              \
    R4(n), R4(n + 2 * 4), R4(n + 1 * 4), R4(n + 3 * 4)

// Lookup table that store the reverse of each table

unsigned int lookuptable[256]
    = { R6(0), R6(2), R6(1), R6(3) };

/* Function to reverse bits of num */

int reverseBits(unsigned int num)
{
    int reverse_num = 0; // Reverse and then rearrange
    // first chunk of 8 bits from right
    reverse_num = lookuptable[num & 0xff] << 24 |
                  // second chunk of 8 bits from right
                  lookuptable[(num >> 8) & 0xff] << 16
                  | lookuptable[(num >> 16) & 0xff] << 8
                  | lookuptable[(num >> 24) & 0xff];
    return reverse_num;
} // driver program to test above function
int main()
{
    int x = 12456;
    printf("%u", reverseBits(x));
    return 0;
}

// Java program to reverse bits using lookup table.
import java.util.*;

class GFG {

    // Lookup table that store the reverse of each table
    public static ArrayList<Integer> lookuptable
        = new ArrayList<Integer>();

    // Generate a lookup table for 32bit operating system
    // using macro
    public static void R2(int n)
    {
        lookuptable.add(n);
        lookuptable.add(n + 2 * 64);
        lookuptable.add(n + 1 * 64);
        lookuptable.add(n + 3 * 64);
    }

    public static void R4(int n)
    {
        R2(n);
        R2(n + 2 * 16);
        R2(n + 1 * 16);
        R2(n + 3 * 16);
    }

    public static void R6(int n)
    {
        R4(n);
        R4(n + 2 * 4);
        R4(n + 1 * 4);
        R4(n + 3 * 4);
    }

    // Function to reverse bits of num
    public static int reverseBits(int num)
    {

        int reverse_num = 0;

        // Reverse and then rearrange

        // first chunk of 8 bits from right
        reverse_num
            = lookuptable.get(num & 0xff) << 24 |
              // second chunk of 8 bits from right
              lookuptable.get((num >> 8) & 0xff) << 16
              | lookuptable.get((num >> 16) & 0xff) << 8
              | lookuptable.get((num >> 24) & 0xff);
        return reverse_num;
    }

    // Driver code
    public static void main(String[] args)
    {
        R6(0);
        R6(2);
        R6(1);
        R6(3);
        // Driver program to test above function
        int x = 12456;
        System.out.println(reverseBits(x));
    }
}

// This code is contributed by phasing17

#Python3 program to reverse bits using lookup table.

# Lookup table that store the reverse of each table
lookuptable = []
 
# Generate a lookup table for 32bit operating system 
# using macro 
def R2(n):
    return lookuptable.extend([n,     n + 2*64,     n + 1*64,     n + 3*64])

def R4(n):
    return R2(n), R2(n + 2*16), R2(n + 1*16), R2(n + 3*16)

def R6(n):
    return R4(n), R4(n + 2*4 ), R4(n + 1*4 ), R4(n + 3*4 )

lookuptable.extend([R6(0), R6(2), R6(1), R6(3)])

# Function to reverse bits of num 
def reverseBits(num):

    reverse_num = 0

    # Reverse and then rearrange 

    # first chunk of 8 bits from right
    reverse_num = lookuptable[num & 0xff ]<<24  | lookuptable[ (num >> 8) & 0xff ]<<16 |lookuptable[ (num >> 16 )& 0xff ]<< 8 | lookuptable[ (num >>24 ) & 0xff ]
  
    return reverse_num

# Driver program to test above function 
x = 12456
print(reverseBits(x))


#This code is contributed by phasing17

// JavaScript program to reverse bits using lookup table.

// Lookup table that stores the reverse of each table.
let lookuptable = [];

// Generate a lookup table for 32-bit operating system using macro.
function R2(n) {
  return lookuptable.push(n, n + 2 * 64, n + 1 * 64, n + 3 * 64);
}

function R4(n) {
  return [R2(n), R2(n + 2 * 16), R2(n + 1 * 16), R2(n + 3 * 16)];
}

function R6(n) {
  return [R4(n), R4(n + 2 * 4), R4(n + 1 * 4), R4(n + 3 * 4)];
}

lookuptable.push(R6(0), R6(2), R6(1), R6(3));

// Function to reverse bits of num.
function reverseBits(num) {
  let reverse_num = 0;

  // Reverse and then rearrange.

  // First chunk of 8 bits from right.
  reverse_num =
    lookuptable[num & 0xff] << 24 |
    lookuptable[(num >> 8) & 0xff] << 16 |
    lookuptable[(num >> 16) & 0xff] << 8 |
    lookuptable[(num >> 24) & 0xff];

  return reverse_num;
}

// Driver program to test above function.
let x = 12456;
console.log(reverseBits(x));

// C# program to reverse bits using lookup table.

using System;
using System.Collections.Generic;

class GFG {

  // Lookup table that store the reverse of each table
  public static List<int> lookuptable = new List<int>();

  // Generate a lookup table for 32bit operating system
  // using macro
  public static void R2(int n)
  {
    lookuptable.Add(n);
    lookuptable.Add(n + 2 * 64);
    lookuptable.Add(n + 1 * 64);
    lookuptable.Add(n + 3 * 64);
  }

  public static void R4(int n)
  {
    R2(n);
    R2(n + 2 * 16);
    R2(n + 1 * 16);
    R2(n + 3 * 16);
  }

  public static void R6(int n)
  {
    R4(n);
    R4(n + 2 * 4);
    R4(n + 1 * 4);
    R4(n + 3 * 4);
  }

  // Function to reverse bits of num
  public static int reverseBits(int num)
  {

    int reverse_num = 0;

    // Reverse and then rearrange

    // first chunk of 8 bits from right
    reverse_num = lookuptable[num & 0xff] << 24 |
      // second chunk of 8 bits from right
      lookuptable[(num >> 8) & 0xff] << 16
      | lookuptable[(num >> 16) & 0xff] << 8
      | lookuptable[(num >> 24) & 0xff];
    return reverse_num;
  }

  // Driver code
  public static void Main(string[] args)
  {
    R6(0);
    R6(2);
    R6(1);
    R6(3);
    int x = 12456;

    // Function call
    Console.WriteLine(reverseBits(x));
  }
}

// This code is contributed by phasing17

Output:

353107968

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