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 nint 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 ndef 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 nfunction 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 conditionint 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 codeint 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 bitsimport 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 conditiondef 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 coden = 15print(countValues(n))# This code is contributed by akashish__ |
C#
// C# code to implement the approachusing 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 conditionfunction 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
1
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 2bool 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 2def 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 2static 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 2function 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 codeimport 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 Codenumber = 0b011print(number)# This code is contributed by akashish__ |
C#
// Conversion into Binary codeusing 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 codelet number = 0b011;console.log(number); |
Output
3
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 = 5b = 7print("Before Swapping, a = ",a," ","b = ",b)a ^= bb ^= aa ^= bprint("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 patternbool 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 patterndef 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 patternfunction 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.
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