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Bits manipulation (Important tactics)

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  • Difficulty Level : Medium
  • Last Updated : 10 Aug, 2022
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Prerequisites: Bitwise operators in C, Bitwise Hacks for Competitive Programming, Bit Tricks for Competitive Programming

Table of Contents

 

1. Compute XOR from 1 to n (direct method):

The  problem can be solved based on the following observations:

Say x = n%4. The XOR value depends on the value if x. If

  • x = 0, then the answer is n.
  • x = 1, then answer is 1.
  • x = 2, then answer is n+1.
  • x = 3, then answer is 0.

Below is the implementation of the above approach.

CPP




// Direct XOR of all numbers from 1 to n
int computeXOR(int n)
{
    if (n % 4 == 0)
        return n;
    if (n % 4 == 1)
        return 1;
    if (n % 4 == 2)
        return n + 1;
    else
        return 0;
}


Java




/*package whatever //do not write package name here */
import java.io.*;
  
class GFG
{
    
  // Direct XOR of all numbers from 1 to n
  public static int computeXOR(int n)
  {
    if (n % 4 == 0)
      return n;
    if (n % 4 == 1)
      return 1;
    if (n % 4 == 2)
      return n + 1;
    else
      return 0;
  }
  
  public static void main (String[] args) {
  
  }
}
  
// This code is contributed by akashish__


Python3




# Direct XOR of all numbers from 1 to n
def computeXOR(n):
    if (n % 4 is 0):
        return n
    if (n % 4 is 1):
        return 1
    if (n % 4 is 2):
        return n + 1
    else:
        return 0
  
        
# This code is contributed by akashish__


C#




using System;
public class GFG
{
  
  // Direct XOR of all numbers from 1 to n
  public static int computeXOR(int n)
  {
  
    if (n % 4 == 0)
  
      return n;
  
    if (n % 4 == 1)
  
      return 1;
  
    if (n % 4 == 2)
  
      return n + 1;
  
    else
  
      return 0;
  
  }
  public static void Main(){}
  
  
}
  
// This code is contributed by akashish__


Javascript




<script>
  
// Direct XOR of all numbers from 1 to n
function computeXOR(n)
{
    if (n % 4 == 0)
        return n;
    if (n % 4 == 1)
        return 1;
    if (n % 4 == 2)
        return n + 1;
    else
        return 0;
}
  
// This code is contributed by Shubham Singh
  
</script>


Input: 6
Output: 7

Refer Compute XOR from 1 to n for details.

2. Count of numbers (x) smaller than or equal to n such that n+x = n^x:

The count of such numbers x can be counted using the following mathematical trick. 

The count = pow(2, count of zero bits).

Refer Equal Sum and XOR for details.

3. How to know if a number is a power of 2?

This can be solved based on the following fact:

If a number N is a power of 2, then the bitwise AND of N and N-1 will be 0. But this will not work if N is 0. So just check these two conditions, if any of these two conditions is true.

Refer check if a number is power of two for details.

Below is the implementation of the above approach.

CPP




//  Function to check if x is power of 2
bool isPowerOfTwo(int x)
{
     // First x in the below expression is
     // for  the case when x is 0 
     return x && (!(x & (x - 1)));
}


Python3




#  Function to check if x is power of 2
def isPowerOfTwo(x):
    
  # First x in the below expression is
  # for  the case when x is 0 
  return x and (not(x & (x - 1)))
  
# This code is contributed by akashish__


C#




using System;
  
public class GFG{
    
  //  Function to check if x is power of 2
static public bool isPowerOfTwo(int x)
{
     // First x in the below expression is
     // for  the case when x is 0 
       return (x != 0) && ((x & (x - 1)) == 0);
}
  
    static public void Main (){
  
    }
}
  
// This code is contributed by akashish__


4. Find XOR of all subsets of a set

We can do it in O(1) time. The answer is always 0 if the given set has more than one element. For sets with a single element, the answer is the value of the single element. 

Refer XOR of the XOR’s of all subsets for details.

5. Find the number of leading, trailing zeroes and number of 1’s

We can quickly find the number of leading, trailing zeroes and number of 1’s in a binary code of an integer in C++ using GCC. 

It can be done by using inbuilt functions i.e.

Number of leading zeroes: __builtin_clz(x)
Number of trailing zeroes : __builtin_ctz(x)
Number of 1-bits: __builtin_popcount(x) 

Refer GCC inbuilt functions for details.

6. Convert binary code directly into an integer in C++

CPP




// Conversion into Binary code
  
#include <iostream>
using namespace std;
  
int main()
{
    auto number = 0b011;
    cout << number;
    return 0;
}


Output

3

7. The Quickest way to swap two numbers:

Two numbers can be swapped easily using the following bitwise operations:

a ^= b;
b ^= a; 
a ^= b;

Refer swap two numbers for more details.  

9. Finding the most significant set bit (MSB):

We can find the most significant set bit in O(1) time for a fixed size integer. For example below code is for 32-bit integer.  

C++




int setBitNumber(int n)
{
    // Below steps set bits after
    // MSB (including MSB)
  
    // Suppose n is 273 (binary
    // is 100010001). It does following
    // 100010001 | 010001000 = 110011001
    n |= n >> 1;
  
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    n |= n >> 2;
  
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
  
    // Increment n by 1 so that
    // there is only one set bit
    // which is just before original
    // MSB. n now becomes 1000000000
    n = n + 1;
  
    // Return original MSB after shifting.
    // n now becomes 100000000
    return (n >> 1);
}


Refer Find most significant set bit of a number for details. 

10. Check if a number has bits in an alternate pattern

We can quickly check if bits in a number are in an alternate pattern (like 101010). 

Compute bitwise XOR (XOR denoted using ^) of n and (n >> 1). If n has an alternate pattern, then n ^ (n >> 1) operation will produce a number having all bits set.

Below is the implementation of the above approach.

C++




// Function to check if a number 
// has bits in alternate pattern
bool bitsAreInAltOrder(unsigned int n)
{
    unsigned int num = n ^ (n >> 1);
      
    // To check if all bits are set in 'num'
    return allBitsAreSet(num);
}


Refer check if a number has bits in alternate pattern for details.

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This article is contributed by Sanchit Garg 1. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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