Data Modelling & AI Data Structure & Algorithm

Bits manipulation (Important tactics)

20 June 2025

38

Prerequisites: Bitwise operators in C, Bitwise Hacks for Competitive Programming, Bit Tricks for Competitive Programming

Table of Contents

Compute XOR from 1 to n (direct method)
Count of numbers (x) smaller than or equal to n such that n+x = n^x
How to know if a number is a power of 2?
Find XOR of all subsets of a set
Find the number of leading, trailing zeroes and number of 1’s
Convert binary code directly into an integer in C++
The Quickest way to swap two numbers
Simple approach to flip the bits of a number
Finding the most significant set bit (MSB)
Check if a number has bits in an alternate pattern

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>

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

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).

C++

// Count of numbers (x) smaller than or equal to n such that n+x = n^x:
// here unset bits means zero bits
#include <bits/stdc++.h>
using namespace std;
 
// function to count number of values less than
// equal to n that satisfy the given condition
int countValues(int n)
{
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    int unset_bits=0;
    while (n)
    {
        if ((n & 1) == 0)
            unset_bits++;
        n=n>>1;
    }
 
    // Return 2 ^ unset_bits i.e. pow(2,count of zero bits)
    return 1 << unset_bits;
}
 
// Driver code
int main()
{
    int n = 15;
    cout << countValues(n);
    return 0;
}
 
// contributed by akashish__

Java

// Count of numbers (x) smaller than or equal to n such that n+x = n^x:
// here unset bits means zero bits
import java.io.*;
class GFG
{
 
  // function to count number of values less than
  // equal to n that satisfy the given condition
  public static int countValues(int n)
  {
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    int unset_bits = 0;
    while (n > 0)
    {
      if ((n & 1) == 0)
        unset_bits++;
      n = n>>1;
    }
 
    // Return 2 ^ unset_bits i.e. pow(2,count of zero bits)
    return 1 << unset_bits;
  }
 
  // Driver code
  public static void main (String[] args) {
    int n = 15;
    System.out.print(countValues(n));
  }
}
 
// This code is contributed by poojaagrawal2.

Python3

# Count of numbers (x) smaller than or equal to n such that n+x = n^x:
# here unset bits means zero bits
 
# function to count number of values less than
# equal to n that satisfy the given condition
def countValues(n):
   
    # unset_bits keeps track of count of un-set
    # bits in binary representation of n
    unset_bits=0
    while (n):
        if ((n & 1) == 0):
            unset_bits+=1
        n=n>>1
 
    # Return 2 ^ unset_bits i.e. pow(2,count of zero bits)
    return 1 << unset_bits
 
# Driver code
n = 15
print(countValues(n))
 
# This code is contributed by akashish__

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Count of numbers (x) smaller than or equal to n such that n+x = n^x:
  // here unset bits means zero bits
 
  // function to count number of values less than
  // equal to n that satisfy the given condition
  static int countValues(int n)
  {
 
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    int unset_bits = 0;
    while (n > 0)
    {
      if ((n & 1) == 0)
        unset_bits++;
      n = n>>1;
    }
 
    // Return 2 ^ unset_bits i.e. pow(2,count of zero bits)
    return 1 << unset_bits;
  }
 
  // Driver code
  public static void Main()
  {
    int n = 15;
    Console.Write(countValues(n));
  }
}
 
// This code is contributed by agrawalpoojaa976.

Javascript

// Count of numbers (x) smaller than or equal to n such that n+x = n^x:
// here unset bits means zero bits
 
// function to count number of values less than
// equal to n that satisfy the given condition
function countValues(n)
{
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    let unset_bits=0;
    while (n)
    {
        if ((n & 1) == 0)
            unset_bits++;
        n=n>>1;
    }
 
    // Return 2 ^ unset_bits i.e. pow(2,count of zero bits)
    return 1 << unset_bits;
}
 
// Driver code
    let n = 15;
    document.write(countValues(n));

Output

Refer Equal Sum and XOR for details.

Time complexity: O(log n)

Auxiliary Space: O(1)

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)));
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  //  Function to check if x is power of 2
  static public boolean isPowerOfTwo(int x)
  {
 
    // First x in the below expression is
    // for  the case when x is 0 
    return (x != 0) && ((x & (x - 1)) == 0);
  }
 
  public static void main (String[] args) {
 
 
  }
}
// contributed by akashish__

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__

Javascript

//  Function to check if x is power of 2
function isPowerOfTwo(x)
{
     // First x in the below expression is
     // for  the case when x is 0 
     return x && (!(x & (x - 1)));
}
 
// This code is contributed by akashish__

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

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;
}

Java

/*package whatever //do not write package name here */
// Conversion into Binary code
import java.io.*;
 
class GFG {
    public static void main(String[] args)
    {
        int number = 0b011;
        System.out.println(number);
    }
}
 
// This code is contributed by akashish__

Python3

# Python Code
number = 0b011
print(number)
 
# This code is contributed by akashish__

C#

// Conversion into Binary code
 
using System;
 
public class GFG {
 
    static public void Main()
    {
 
        // Code
        int number = 0b011;
        Console.WriteLine(number);
    }
}
 
// This code is contributed by karthik

Javascript

// Conversion into Binary code
 
let number = 0b011;
console.log(number);

Output

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

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;

C++

#include <iostream>
using namespace std;
 
int main()
{
    int a = 5;
      int b = 7;
      cout<<"Before Swapping, a = "<<a<<" "<<"b = "<<b<<endl;
    a ^= b;
    b ^= a;
    a ^= b;
      cout<<"After Swapping, a = "<<a<<" "<<"b = "<<b<<endl;
    return 0;
}
 
// This code is contributed by akashish__

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
 
class GFG {
  public static void main(String[] args)
  {
    int a = 5;
    int b = 7;
    System.out.print("Before Swapping, a = ");
    System.out.print(a);
    System.out.print(" ");
    System.out.print("b = ");
    System.out.print(b);
    System.out.println("");
    a ^= b;
    b ^= a;
    a ^= b;
    System.out.print("After Swapping, a = ");
    System.out.print(a);
    System.out.print(" ");
    System.out.print("b = ");
    System.out.print(b);
  }
}
 
// This code is contributed by akashish__

Python3

a = 5
b = 7
print("Before Swapping, a = ",a," ","b = ",b)
a ^= b
b ^= a
a ^= b
print("After Swapping, a = ",a," ","b = ",b)
 
# This code is contributed by akashish__

C#

using System;
 
public class GFG {
 
    static public void Main()
    {
 
        int a = 5;
        int b = 7;
        Console.WriteLine("Before Swapping, a = " + a + " "
                          + "b = " + b);
        a ^= b;
        b ^= a;
        a ^= b;
        Console.WriteLine("After Swapping, a = " + a + " "
                          + "b = " + b);
    }
}
// This code is contributed by akashish__

Javascript

let a = 5;
let b = 7;
console.log("Before Swapping, a = " , a , " " , "b = " , b);
a ^= b;
b ^= a;
a ^= b;
console.log("After Swapping, a = " , a , " " , "b = " , b);
 
// This code is contributed by akashish__

Output

Before Swapping, a = 5 b = 7
After Swapping, a = 7 b = 5

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

Refer swap two numbers for more details.

8. 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);
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
 
  public static 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);
  }
 
 
  public static void main (String[] args) {
  }
}
 
// This code is contributed by akashish__

Python3

def setBitNumber(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)
   
# This code is contributed by akashish__

C#

using System;
 
public class GFG{
   
  public static 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);
}
 
    static public void Main (){
 
        // Code
    }
}
 
// This code is contributed by akashish__

Javascript

function setBitNumber(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);
}
 
// This code is contributed by akashish__

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

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

9. 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 all the bits
// are set or not in the binary
// representation of 'n'
static bool allBitsAreSet(int n)
{
    // if true, then all bits are set
    if (((n + 1) & n) == 0)
        return true;
 
    // else all bits are not set
    return false;
}
 
// 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);
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  // function to check if all the bits
  // are set or not in the binary
  // representation of 'n'
  public static boolean allBitsAreSet(long n)
  {
 
    // if true, then all bits are set
    if (((n + 1) & n) == 0)
      return true;
 
    // else all bits are not set
    return false;
  }
 
  // Function to check if a number
  // has bits in alternate pattern
  public static boolean bitsAreInAltOrder(long n)
  {
    long num = n ^ (n >> 1);
 
    // To check if all bits are set in 'num'
    return allBitsAreSet(num);
  }
  public static void main (String[] args) {
 
  }
}
 
// This code is contributed by akashish__

Python3

# function to check if all the bits
# are set or not in the binary
# representation of 'n'
def allBitsAreSet(n):
  # if true, then all bits are set
  if (((n + 1) & n) == 0):
    return True
 
  # else all bits are not set
  return False
 
# Function to check if a number
# has bits in alternate pattern
def bitsAreInAltOrder(n):
  num = n ^ (n >> 1)
 
  # To check if all bits are set in 'num'
  return allBitsAreSet(num)
 
 
# This code is contributed by akashish__

C#

using System;
public class GFG 
{
 
    // function to check if all the bits
    // are set or not in the binary
    // representation of 'n'
    public static bool allBitsAreSet(uint n)
    {
       
        // if true, then all bits are set
        if (((n + 1) & n) == 0)
            return true;
 
        // else all bits are not set
        return false;
    }
 
    // Function to check if a number
    // has bits in alternate pattern
    public static bool bitsAreInAltOrder(uint n)
    {
        uint num = n ^ (n >> 1);
 
        // To check if all bits are set in 'num'
        return allBitsAreSet(num);
    }
 
    static public void Main() {}
}
 
// This code is contributed by akashish__

Javascript

// function to check if all the bits
// are set or not in the binary
// representation of 'n'
function allBitsAreSet(n)
{
    // if true, then all bits are set
    if (((n + 1) & n) == 0)
        return true;
 
    // else all bits are not set
    return false;
}
 
// Function to check if a number
// has bits in alternate pattern
function bitsAreInAltOrder(n)
{
    let num = n ^ (n >> 1);
 
    // To check if all bits are set in 'num'
    return allBitsAreSet(num);
}
 
// This code is contributed by akashish__

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

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

DSA Self Paced Course

This article is contributed by Sanchit Garg 1. If you like neveropen 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 neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

1 COMMENT

All about Bit Manipulation – Neveropen Tech 12 August 2025 At 6:09 am

[…] Bit Manipulation is a technique used in a variety of problems to get the solution in an optimized way. This technique is very effective from a Competitive Programming point of view. It is all about Bitwise Operators which directly works upon binary numbers or bits of numbers that help the implementation fast. Below are the Bitwise Operators that are used: […]

Log in to leave a comment

Bits manipulation (Important tactics)

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

CPP

Java

Python3

C#

Javascript

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

C++

Java

Python3

C#

Javascript

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

CPP

Java

Python3

C#

Javascript

4. Find XOR of all subsets of a set

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

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

CPP

Java

Python3

C#

Javascript

7. The Quickest way to swap two numbers:

C++

Java

Python3

C#

Javascript

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

C++

Java

Python3

C#

Javascript

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

C++

Java

Python3

C#

Javascript

1 COMMENT

LEAVE A REPLY Cancel reply

Most Popular

EDITOR PICKS

POPULAR POSTS

POPULAR CATEGORY

ABOUT US

FOLLOW US