Set all odd bits of a number
Last Updated :
26 Apr, 2023
Given a number, the task is to set all odd bits of a number. Positions of bits are counted from LSB (least significant bit) to MSB (Most significant bit). Position of LSB is considered as 1.
Examples :
Input : 20
Output : 21
Explanation : Binary representation of 20
is 10100. Setting all odd
bits make the number 10101 which is binary
representation of 21.
Input : 10
Output : 15
Method 1 (Using OR)
1. First generate a number that contains odd position bits.
2. Take OR with the original number. Note that 1 | 1 = 1 and 1 | 0 = 1.
Let’s understand this approach with below code.
C++
// CPP code Set all odd bits
// of a number
#include <iostream>
using namespace std;
// set all odd bit
int oddbitsetnumber(int n)
{
int count = 0;
// res for store 010101.. number
int res = 0;
// generate number form of 010101.....till
// temp size
for (int temp = n; temp > 0; temp >>= 1) {
// if bit is odd, then generate
// number and or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
return (n | res);
}
// Driver code
int main()
{
int n = 10;
cout << oddbitsetnumber(n);
return 0;
}
// Java code to Set all odd
// bits of a number
class GFG
{
// set all odd bit
static int oddbitsetnumber(int n)
{
int count = 0;
// res for store 010101.. number
int res = 0;
// generate number form of
// 010101.....till temp size
for (int temp = n; temp > 0; temp >>= 1)
{
// if bit is odd, then generate
// number and or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
return (n | res);
}
// Driver code
public static void main(String[] args)
{
int n = 10;
System.out.println(oddbitsetnumber(n));
}
}
// This code is contributed
// by prerna saini
''' Python3 code Set all odd bits
of a number'''
# set all odd bit
def oddbitsetnumber(n):
count = 0
# res for store 010101.. number
res = 0
# generate number form of 010101.....till
# temp size
temp = n
while temp > 0:
# if bit is odd, then generate
# number and or with res
if count % 2 == 0:
res |= (1 << count)
count += 1
temp >>= 1
return (n | res)
n = 10
print (oddbitsetnumber(n))
#This code is contributed by Shreyanshi Arun.
// C# code to Set all odd
// bits of a number
using System;
class GFG
{
static int oddbitsetnumber(int n)
{
int count = 0;
// res for store 010101.. number
int res = 0;
// generate number form of
// 010101.....till temp size
for (int temp = n; temp > 0;
temp >>= 1)
{
// if bit is odd, then
// generate number and
// or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
return (n | res);
}
// Driver Code
static public void Main ()
{
int n = 10;
Console.WriteLine(oddbitsetnumber(n));
}
}
// This code is contributed
// by prerna ajit
<?php
// php code Set all odd
// bits of a number
// set all odd bit
function oddbitsetnumber($n)
{
$count = 0;
// res for store 010101..
// number
$res = 0;
// generate number form of
// 010101... till temp size
for ($temp = $n; $temp > 0; $temp >>= 1)
{
// if bit is odd, then generate
// number and or with res
if ($count % 2 == 0)
$res |= (1 << $count);
$count++;
}
return ($n | $res);
}
// Driver code
$n = 10;
echo oddbitsetnumber($n);
// This code is contributed by mits
?>
<script>
// JavaScript Program to Set all odd
// bits of a number
// set all odd bit
function oddbitsetnumber(n)
{
let count = 0;
// res for store 010101.. number
let res = 0;
// generate number form of
// 010101.....till temp size
for (let temp = n; temp > 0; temp >>= 1)
{
// if bit is odd, then generate
// number and or with res
if (count % 2 == 0)
res |= (1 << count);
count++;
}
return (n | res);
}
// Driver Code
let n = 10;
document.write(oddbitsetnumber(n));
// This code is contributed by chinmoy1997pal.
</script>
Method 2 (A O(1) solution for 32 bit numbers)
C++
// Efficient CPP program to set all
// odd bits number
#include <iostream>
using namespace std;
// return MSB set number
int getmsb(int n)
{
// set all bits including MSB.
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// return MSB
return (n + 1) >> 1;
}
// Returns a number of same size (MSB at
// same position) as n and all odd bits
// set.
int getevenbits(int n)
{
n = getmsb(n);
// generate odd bits like 010101..
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// if bits is even then shift by 1
if ((n&1) == 0)
n = n >> 1;
// return odd set bits number
return n;
}
// set all odd bits here
int setalloddbits(int n)
{
// take OR with odd set bits number
return n | getevenbits(n);
}
// Driver code
int main()
{
int n = 10;
cout << setalloddbits(n);
return 0;
}
// Efficient Java program to set
// all odd bits number
class GFG {
// return MSB set number
static int getmsb(int n)
{
// set all bits including MSB.
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// return MSB
return (n + 1) >> 1;
}
// Returns a number of same
// size (MSB at same position)
// as n and all odd bits set.
static int getevenbits(int n)
{
n = getmsb(n);
// generate odd bits
// like 010101..
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// if bits is even
// then shift by 1
if ((n & 1) == 0)
n = n >> 1;
// return odd set bits number
return n;
}
// set all odd bits here
static int setalloddbits(int n)
{
// take OR with odd
// set bits number
return n | getevenbits(n);
}
// Driver code
public static void main(String[] args)
{
int n = 10;
System.out.println(setalloddbits(n));
}
}
// This code is contributed
// by prerna saini
# Efficient python3 program to
# set all odd bits number
import math
# return MSB set number
def getmsb( n):
# set all bits including MSB.
n |= n >> 1
n |= n >> 2
n |= n >> 4
n |= n >> 8
n |= n >> 16
# return MSB
return (n + 1) >> 1
# Returns a number of same
# size (MSB at same position)
# as n and all odd bits set.
def getevenbits(n):
n = getmsb(n)
# generate odd bits
# like 010101..
n |= n >> 2
n |= n >> 4
n |= n >> 8
n |= n >> 16
# if bits is even
# then shift by 1
if ((n & 1) == 0):
n = n >> 1
# return odd set bits number
return n
# set all odd bits here
def setalloddbits( n):
# take OR with odd
# set bits number
return n | getevenbits(n)
# Driver Program
n = 10
print(setalloddbits(n))
# This code is contributed
# by Gitanjali.
// Efficient C# program to
// set all odd bits number
using System;
class GFG
{
// return MSB set number
static int getmsb(int n)
{
// set all bits
// including MSB.
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// return MSB
return (n + 1) >> 1;
}
// Returns a number of same
// size (MSB at same position)
// as n and all odd bits set.
static int getevenbits(int n)
{
n = getmsb(n);
// generate odd bits
// like 010101..
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// if bits is even
// then shift by 1
if ((n & 1) == 0)
n = n >> 1;
// return odd
// set bits number
return n;
}
// set all odd bits here
static int setalloddbits(int n)
{
// take OR with odd
// set bits number
return n | getevenbits(n);
}
// Driver Code
static public void Main ()
{
int n = 10;
Console.WriteLine(setalloddbits(n));
}
}
// This code is contributed ajit
<?php
// Efficient php program to
// set all odd bits number
// return MSB set number
function getmsb($n)
{
// set all bits
// including MSB.
$n |= $n >> 1;
$n |= $n >> 2;
$n |= $n >> 4;
$n |= $n >> 8;
$n |= $n >> 16;
// return MSB
return ($n + 1) >> 1;
}
// Returns a number of
// same size (MSB at
// same position) as n
// and all odd bits set
function getevenbits($n)
{
$n = getmsb($n);
// generate odd bits
// like 010101..
$n |= $n >> 2;
$n |= $n >> 4;
$n |= $n >> 8;
$n |= $n >> 16;
// if bits is even
// then shift by 1
if (($n&1) == 0)
$n = $n >> 1;
// return odd set
// bits number
return $n;
}
// set all odd bits here
function setalloddbits($n)
{
// take OR with odd
// set bits number
return $n | getevenbits($n);
}
// Driver code
$n = 10;
echo setalloddbits($n);
// This code is contributed by mits
?>
// Efficient Javascript program to
// set all odd bits number
// Return MSB set number
function getmsb(n)
{
// Set all bits
// including MSB.
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// Return MSB
return (n + 1) >> 1;
}
// Returns a number of same
// size (MSB at same position)
// as n and all odd bits set.
function getevenbits(n)
{
n = getmsb(n);
// Generate odd bits
// like 010101..
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
// If bits is even
// then shift by 1
if ((n & 1) == 0)
n = n >> 1;
// Return odd
// set bits number
return n;
}
// Set all odd bits here
function setalloddbits(n)
{
// Take OR with odd
// set bits number
return n | getevenbits(n);
}
// Driver Code
let n = 10;
console.log(setalloddbits(n));
// This code is contributed by decode2207
Another approach:
O(1) approach which supports even long long (64-bit).
C++
#include <iostream>
using namespace std;
long long int setAllOddBits(long long int n) {
// code here
long long mask;
//find required num such that it
//is of form 000.....1010
if(__builtin_clz(n)%2==0){
mask=0x5555555555555555>>(__builtin_clzll(n));
}else{
mask=0xAAAAAAAAAAAAAAAA>>(__builtin_clzll(n));
}
// return toggled number
return n|mask;
}
// Driver code
int main()
{
int n = 10;
cout << setAllOddBits(n);
return 0;
}
// Java equivalent code for the given C++ code
// with the same comments
public class Main {
public static long setAllOddBits(long n)
{
// find required num such that it is of form
// 000.....1010
long mask;
if (Integer.numberOfLeadingZeros((int)n) % 2 == 0) {
mask = 0x5555555555555555L
>>> Long.numberOfLeadingZeros(n);
}
else {
mask = 0xAAAAAAAAAAAAAAAAL
>>> Long.numberOfLeadingZeros(n);
} // return toggled number
return n | mask;
}
public static void main(String[] args)
{
int n = 10;
System.out.println(setAllOddBits(n));
}
}
# Python 3 implementation
def setAllOddBits(n):
# find required num such that it is of form
# 000.....1010
if (bin(n).count('1') % 2 == 0):
mask = 0x5555555555555555 >> (64 - len(bin(n)) + 2)
else:
mask = 0xAAAAAAAAAAAAAAAA >> (64 - len(bin(n)) + 2)
# return toggled number
return n | mask
# Driver code
n = 10
print(setAllOddBits(n))
# This code is contributed by phasing17.
using System;
public class Program {
public static long SetAllOddBits(long n)
{
ulong un = (ulong)n;
ulong mask;
// find required num such that it
// is of form 000.....1010
if (CountLeadingZeros(un) % 2 == 0) {
mask = 0x5555555555555555UL
>> CountLeadingZeros(un);
}
else {
mask = 0xAAAAAAAAAAAAAAAAL
>> CountLeadingZeros(un);
}
// return toggled number
return (long)(un | mask);
}
public static int CountLeadingZeros(ulong x)
{
int count = 0;
while (x != 0) {
x >>= 1;
count++;
}
return 64 - count;
}
// Driver code
public static void Main()
{
int n = 10;
Console.WriteLine(SetAllOddBits(n));
}
}
// Javascript code addition
function setAllOddBits(n) {
// find required num such that it is of form
// 000.....1010
let x = 17;
if (n.toString(2).split('1').length % 2 === 0) {
mask = 0x5555555555555555 >> (64 - n.toString(2).length + 2);
} else {
mask = 0xAAAAAAAAAAAAAAAA >> (64 - n.toString(2).length + 2);
}
// return toggled number
return n | mask + x;
}
let n = 10;
console.log(setAllOddBits(n));
// The code is contributed by Nidhi goel.
Time complexity: O(1)
Space complexity: O(1)
Another Approach:
To set a bit, we can take OR of 1 and that bit (as 1 | 1 = 1, and 1 | 0 = 1). Therefore, to set all odd bits of an n bit number, we need to use a bit mask which is an n bit binary number with all odd bits set. This mask can be generated in 1 step using the formula of sum of a geometric progression, as an n bit number ...0101 is equal to 20 + 22 + 24 + .... 2 (n - 1 - !(n % 1)) .
C++
#include <bits/stdc++.h>
using namespace std;
// function to set all the odd bits
long long int setAllOddBits(long long int n)
{
// calculating number of bits using log
int numOfBits = 1 + (int)log2(n);
// calculating the max power of GP series
int m = (numOfBits + (numOfBits % 2)) / 2;
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 1, r = 4, n = m
int mask = (pow(4, m) - 1) / 3;
// setting all odd bits using mask | n
return mask | n;
}
// Driver code
int main()
{
int n = 10;
// function call
cout << setAllOddBits(n);
return 0;
}
// this code is contributed by phasing17
// Java code to implement the approach
class GFG {
// function to set all the odd bits
static int setAllOddBits(int n)
{
// calculating number of bits using log
int numOfBits
= 1 + (int)(Math.log(n) / Math.log(2));
// calculating the max power of GP series
int m = (numOfBits + (numOfBits % 2)) / 2;
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 1, r = 4, n = m
int mask = (int)(Math.pow(4, m) - 1) / 3;
// setting all odd bits using mask | n
return mask | n;
}
// Driver code
public static void main(String[] args)
{
int n = 10;
// function call
System.out.println(setAllOddBits(n));
}
}
// this code is contributed by phasing17
# Python program to implement the approach
import math
# function to set all the odd bits
def setAllOddBits(n):
# calculating number of bits using log
numOfBits = 1 + math.floor(math.log2(n))
# calculating the max power of GP series
m = math.floor((numOfBits + (numOfBits % 2)) / 2)
# calculating mask using GP sum
# which is a(r ^ n - 1) / (r - 1)
# where a = 1, r = 4, n = m
mask = math.floor((math.pow(4, m) - 1) / 3)
# setting all odd bits using mask | n
return mask | n
# Driver code
n = 10
# function call
print(setAllOddBits(n))
using System;
public class Program
{
// function to set all the odd bits
static long setAllOddBits(long n)
{
// calculating number of bits using log
int numOfBits = 1 + (int)Math.Log(n, 2);
// calculating the max power of GP series
int m = (numOfBits + (numOfBits % 2)) / 2;
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 1, r = 4, n = m
int mask = (int)(Math.Pow(4, m) - 1) / 3;
// setting all odd bits using mask | n
return mask | n;
}
// Driver code
public static void Main()
{
int n = 10;
// function call
Console.WriteLine(setAllOddBits(n));
}
}
//JavaScript program to implement the approach
// function to set all the odd bits
function setAllOddBits(n)
{
// calculating number of bits using log
let numOfBits = 1 + Math.floor(Math.log2(n));
// calculating the max power of GP series
let m = Math.floor((numOfBits + (numOfBits % 2)) / 2);
// calculating mask using GP sum
// which is a(r ^ n - 1) / (r - 1)
// where a = 1, r = 4, n = m
let mask = Math.floor((Math.pow(4, m) - 1) / 3);
// setting all odd bits using mask | n
return mask | n;
}
// Driver code
let n = 10;
// function call
console.log(setAllOddBits(n));
// this code is contributed by phasing17
Time complexity: O(1)
Space complexity: O(1)
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