Given an unsigned integer, reverse all bits of it and return the number with reversed bits.
Input : n = 1
Output : 2147483648
Explanation : On a machine with size of unsigned bit as 32. Reverse of 0….001 is 100….0.Input : n = 2147483648
Output : 1
Method1 – Simple: Loop through all the bits of an integer. If a bit at ith position is set in the i/p no. then set the bit at (NO_OF_BITS – 1) – i in o/p. Where NO_OF_BITS is number of bits present in the given number.
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;// Function to reverse bits of numunsigned int reverseBits(unsigned int num){ unsigned int NO_OF_BITS = sizeof(num) * 8; unsigned int reverse_num = 0; int i; for (i = 0; i < NO_OF_BITS; i++) { if ((num & (1 << i))) reverse_num |= 1 << ((NO_OF_BITS - 1) - i); } return reverse_num;}// Driver codeint main(){ unsigned int x = 2; cout << reverseBits(x); return 0;}// This code is contributed By Shivam Tiwari |
c
// C code to implement the approach#include <stdio.h>// Function to reverse bits of numunsigned int reverseBits(unsigned int num){ unsigned int NO_OF_BITS = sizeof(num) * 8; unsigned int reverse_num = 0; int i; for (i = 0; i < NO_OF_BITS; i++) { if ((num & (1 << i))) reverse_num |= 1 << ((NO_OF_BITS - 1) - i); } return reverse_num;}// Driver codeint main(){ unsigned int x = 2; printf("%u", reverseBits(x)); getchar();} |
Python
def reverse_bits(num): NO_OF_BITS = 32 reverse_num = 0 for i in range(NO_OF_BITS): if (num & (1 << i)): reverse_num |= 1 << ((NO_OF_BITS - 1) - i) return reverse_num# Driver codex = 2print(reverse_bits(x)) |
C#
using System;public class GFG { // Function to reverse bits of num public static uint ReverseBits(uint num) { uint NO_OF_BITS = (uint)(sizeof(uint) * 8); uint reverse_num = 0; int i; for (i = 0; i < NO_OF_BITS; i++) { if ((num & (1u << i)) != 0) reverse_num |= 1u << ((int)(NO_OF_BITS - 1) - i); } // Return the reversed number return reverse_num; } // Driver Code public static void Main() { uint x = 2u; Console.WriteLine(ReverseBits(x)); }} |
Javascript
function reverseBits(num) { let NO_OF_BITS = 32; let reverse_num = 0; for (let i = 0; i < NO_OF_BITS; i++) { if ((num & (1 << i)) !== 0) { reverse_num |= 1 << (NO_OF_BITS - 1) - i; } } return reverse_num;}// Driver codelet x = 2;console.log(reverseBits(x)); |
1073741824
Time Complexity: O(Log n). Time complexity would be Log(num) as there are log(num) bits in a binary number “num” and we’re looping through all bits.
Auxiliary space: O(1)
Method 2 – Standard: The idea is to keep putting set bits of the num in reverse_num until num becomes zero. After num becomes zero, shift the remaining bits of reverse_num. Let num is stored using 8 bits and num be 00000110. After the loop you will get reverse_num as 00000011. Now you need to left shift reverse_num 5 more times and you get the exact reverse 01100000.
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;// Function to reverse bits of numunsigned int reverseBits(unsigned int num){ unsigned int count = sizeof(num) * 8 - 1; unsigned int reverse_num = num; num >>= 1; while (num) { reverse_num <<= 1; reverse_num |= num & 1; num >>= 1; count--; } reverse_num <<= count; return reverse_num;}// Driver's codeint main(){ unsigned int x = 1; cout << reverseBits(x) << endl; return 0;}// This code is contributed By Shivam Tiwari |
C
// C code to implement the approach#include <stdio.h>// Function to reverse bits of numunsigned int reverseBits(unsigned int num){ unsigned int count = sizeof(num) * 8 - 1; unsigned int reverse_num = num; num >>= 1; while (num) { reverse_num <<= 1; reverse_num |= num & 1; num >>= 1; count--; } reverse_num <<= count; return reverse_num;}// Driver's codeint main(){ unsigned int x = 1; printf("%u", reverseBits(x)); getchar();} |
Python3
def reverseBits(num): count = 32 - 1 reverse_num = num num >>= 1 while (num): reverse_num <<= 1 reverse_num |= num & 1 num >>= 1 count -= 1 reverse_num <<= count return reverse_num# Driver's codeif __name__ == '__main__': x = 1 print(reverseBits(x))# This code is contributed by shivhack999 |
C#
using System;class Program{ // Function to reverse bits of num static uint ReverseBits(uint num) { uint count = sizeof(uint) * 8 - 1; uint reverse_num = num; num >>= 1; while (num != 0) { reverse_num <<= 1; reverse_num |= num & 1; num >>= 1; count--; } reverse_num <<= (int)count; return reverse_num; } // Driver's code static void Main() { uint x = 1; Console.WriteLine(ReverseBits(x)); }}// This Code is Contributed by Shivam Tiwari |
2147483648
Time Complexity: O(logn) where n is the given number
Auxiliary space: O(1)
Method 3 – Lookup Table: We can reverse the bits of a number in O(1) if we know the size of the number. We can implement it using look up table. Please refer Reverse bits using lookup table in O(1) time for details.
Source : https://graphics.stanford.edu/~seander/bithacks.html
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