Reverse actual bits of the given number

03 Jun, 2022
Given a non-negative integer n. The problem is to reverse the bits of n and print the number obtained after reversing the bits. Note that the actual binary representation of the number is being considered for reversing the bits, no leading 0’s are being considered.Examples :  
Input : 11
Output : 13
(11)10 = (1011)2.
After reversing the bits we get:
(1101)2 = (13)10.

Input : 10
Output : 5
(10)10 = (1010)2.
After reversing the bits we get:
(0101)2 = (101)2
        = (5)10.
 
In this approach, one by one bit in the binary representation of n is being obtained with the help of bitwise right shift operation and they are being accumulated in rev with the help of bitwise left shift operation. Algorithm:  
 
C++
C
Java
Python3
C#
PHP
Javascript
// C++ implementation to reverse bits of a number#include <bits/stdc++.h> using namespace std; // function to reverse bits of a numberunsigned int reverseBits(unsigned int n){    unsigned int rev = 0;     // traversing bits of 'n' from the right    while (n > 0) {        // bitwise left shift        // 'rev' by 1        rev <<= 1;         // if current bit is '1'        if (n & 1 == 1)            rev ^= 1;         // bitwise right shift        // 'n' by 1        n >>= 1;    }     // required number    return rev;} // Driver program to test aboveint main(){    unsigned int n = 11;    cout << reverseBits(n);    return 0;}
// C implementation to reverse bits of a number#include <stdio.h> // function to reverse bits of a numberunsigned int reverseBits(unsigned int n){    unsigned int rev = 0;     // traversing bits of 'n' from the right    while (n > 0) {        // bitwise left shift 'rev' by 1        rev <<= 1;         // if current bit is '1'        if (n & 1 == 1)            rev ^= 1;         // bitwise right shift 'n' by 1        n >>= 1;    }    // required number    return rev;} // Driver program to test aboveint main(){    unsigned int n = 11;    printf("%u",reverseBits(n));    return 0;} // This code is contributed by Sania Kumari Gupta (kriSania804)
// Java implementation to// reverse bits of a numberclass GFG{    // function to reverse bits of a number    public static int reverseBits(int n)    {        int rev = 0;         // traversing bits of 'n'        // from the right        while (n > 0)        {            // bitwise left shift            // 'rev' by 1            rev <<= 1;             // if current bit is '1'            if ((int)(n & 1) == 1)                rev ^= 1;             // bitwise right shift            //'n' by 1            n >>= 1;        }        // required number        return rev;    }     // Driver code    public static void main(String[] args)    {        int n = 11;        System.out.println(reverseBits(n));    }} // This code is contributed// by prerna saini.
# Python 3 implementation to# reverse bits of a number  # function to reverse# bits of a numberdef reverseBits(n) :         rev = 0         # traversing bits of 'n' from the right    while (n > 0) :                 # bitwise left shift 'rev' by 1        rev = rev << 1                 # if current bit is '1'        if (n & 1 == 1) :            rev = rev ^ 1                 # bitwise right shift 'n' by 1        n = n >> 1                  # required number    return rev     # Driver coden = 11print(reverseBits(n))  # This code is contributed# by Nikita Tiwari.
// C# implementation to// reverse bits of a numberusing System;class GFG{    // function to reverse bits of a number    public static int reverseBits(int n)    {        int rev = 0;         // traversing bits of 'n'        // from the right        while (n > 0)        {            // bitwise left shift            // 'rev' by 1            rev <<= 1;             // if current bit is '1'            if ((int)(n & 1) == 1)                rev ^= 1;             // bitwise right shift            //'n' by 1            n >>= 1;        }        // required number        return rev;    }     // Driver code    public static void Main()    {        int n = 11;        Console.WriteLine(reverseBits(n));    }} // This code is contributed// by vt_m.
<?php// PHP implementation to reverse// bits of a number // function to reverse// bits of a numberfunction reverseBits($n){    $rev = 0;         // traversing bits of 'n'    // from the right    while ($n > 0)    {        // bitwise left shift        // 'rev' by 1        $rev <<= 1;                 // if current bit is '1'        if ($n & 1 == 1)            $rev ^= 1;                 // bitwise right shift        // 'n' by 1        $n >>= 1;                 }         // required number    return $rev;} // Driver code$n = 11;echo reverseBits($n); // This code is contributed by mits?>
<script> // JavaScript program  to// reverse bits of a number     // function to reverse bits of a number    function reverseBits(n)    {        let rev = 0;           // traversing bits of 'n'        // from the right        while (n > 0)        {            // bitwise left shift            // 'rev' by 1            rev <<= 1;               // if current bit is '1'            if ((n & 1) == 1)                rev ^= 1;               // bitwise right shift            //'n' by 1            n >>= 1;        }        // required number        return rev;    } // Driver code         let n = 11;        document.write(reverseBits(n)); </script>
13
Time Complexity: O(num), where num is the number of bits in the binary representation of n.
Space Complexity: O(1) 
Mithun Kumar
code_hunt
rohitmishra051000
krisania804
Reverse
Bit Magic
Bit Magic
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