Find most significant set bit of a number
Last Updated :
06 Apr, 2023
Given a number, find the greatest number less than the given a number which is the power of two or find the most significant bit number .
Examples:
Input: 10
Output: 8
Explanation:
Greatest number which is a Power of 2 less than 10 is 8
Binary representation of 10 is 1010
The most significant bit corresponds to decimal number 8.
Input: 18
Output: 16
A simple solution is to one by one divide n by 2 until it becomes 0 and increment a count while doing this. This count actually represents the position of MSB.
C++
// Simple CPP program to find MSB number
// for given POSITIVE n.
#include <iostream>
using namespace std;
int setBitNumber(int n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return (1 << msb);
}
// Driver code
int main()
{
int n = 0;
cout << setBitNumber(n);
n = ~(int)0; // test for possible overflow
cout << "\n" << (unsigned int)setBitNumber(n);
return 0;
}
#include <stdio.h>
int setBitNumber(int n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return (1 << msb);
}
int main() {
int n = 0;
printf("%d \n",setBitNumber(n));
n = ~(int)0; // test for possible overflow
int ns = (unsigned int)setBitNumber(n);
printf("%d",ns);
return 0;
}
// Simple Java program to find
// MSB number for given n.
import java.io.*;
class GFG {
static int setBitNumber(int n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return (1 << msb);
}
// Driver code
public static void main(String[] args)
{
int n = 0;
System.out.println(setBitNumber(n));
}
}
// This code is contributed by ajit
# Simple Python3 program
# to find MSB number
# for given n.
def setBitNumber(n):
if (n == 0):
return 0;
msb = 0;
n = int(n / 2);
while (n > 0):
n = int(n / 2);
msb += 1;
return (1 << msb);
# Driver code
n = 0;
print(setBitNumber(n));
# This code is contributed
# by mits
// Simple C# program to find
// MSB number for given n.
using System;
class GFG {
static int setBitNumber(int n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return (1 << msb);
}
// Driver code
static public void Main()
{
int n = 0;
Console.WriteLine(setBitNumber(n));
}
}
// This code is contributed
// by akt_mit
<?php
// Simple PhP program
// to find MSB number
// for given n.
function setBitNumber($n)
{
if ($n == 0)
return 0;
$msb = 0;
$n = $n / 2;
while ($n != 0)
{
$n = $n / 2;
$msb++;
}
return (1 << $msb);
}
// Driver code
$n = 0;
echo setBitNumber($n);
// This code is contributed
// by akt_mit
?>
<script>
// Javascript program
// to find MSB number
// for given n.
function setBitNumber(n)
{
if (n == 0)
return 0;
let msb = 0;
n = n / 2;
while (n != 0)
{
n = $n / 2;
msb++;
}
return (1 << msb);
}
// Driver code
let n = 0;
document.write (setBitNumber(n));
// This code is contributed by Bobby
</script>
Time Complexity: O(logn)
Auxiliary Space: O(1)
An efficient solution for a fixed size integer (say 32 bits) is to one by one set bits, then add 1 so that only the bit after MSB is set. Finally right shift by 1 and return the answer. This solution does not require any condition checking.
C++
// CPP program to find MSB number for ANY given n.
#include <iostream>
#include <limits.h>
using namespace std;
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;
// The naive approach would increment n by 1,
// so only the MSB+1 bit will be set,
// So now n theoretically becomes 1000000000.
// All the would remain is a single bit right shift:
// n = n + 1;
// return (n >> 1);
//
// ... however, this could overflow the type.
// To avoid overflow, we must retain the value
// of the bit that could overflow:
// n & (1 << ((sizeof(n) * CHAR_BIT)-1))
// and OR its value with the naive approach:
// ((n + 1) >> 1)
n = ((n + 1) >> 1) |
(n & (1 << ((sizeof(n) * CHAR_BIT)-1)));
return n;
}
// Driver code
int main()
{
int n = 273;
cout << setBitNumber(n);
n = ~(int)0; // test for possible overflow
cout << "\n" << (unsigned int)setBitNumber(n);
return 0;
}
#include <stdio.h>
#include <limits.h>
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;
// The naive approach would increment n by 1,
// so only the MSB+1 bit will be set,
// So now n theoretically becomes 1000000000.
// All the would remain is a single bit right shift:
// n = n + 1;
// return (n >> 1);
//
// ... however, this could overflow the type.
// To avoid overflow, we must retain the value
// of the bit that could overflow:
// n & (1 << ((sizeof(n) * CHAR_BIT)-1))
// and OR its value with the naive approach:
// ((n + 1) >> 1)
n = ((n + 1) >> 1) |
(n & (1 << ((sizeof(n) * CHAR_BIT)-1)));
return n;
}
int main() {
int n = 273;
printf("%d\n",setBitNumber(n));
return 0;
}
// Java program to find MSB
// number for given n.
class GFG {
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);
}
// Driver code
public static void main(String arg[])
{
int n = 273;
System.out.print(setBitNumber(n));
}
}
// This code is contributed by Anant Agarwal.
# Python program to find
# MSB number for given n.
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)
# Driver code
n = 273
print(setBitNumber(n))
# This code is contributed
# by Anant Agarwal.
// C# program to find MSB number for given n.
using System;
class GFG {
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);
}
// Driver code
public static void Main()
{
int n = 273;
Console.WriteLine(setBitNumber(n));
}
}
// This code is contributed by Sam007.
<?php
// PHP program to find
// MSB number for given n.
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);
}
// Driver code
$n = 273;
echo setBitNumber($n);
// This code is contributed
// by akt_mit
?>
<script>
// Javascript program to find MSB
// number for given n.
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);
}
// Driver code
let n = 273;
document.write(setBitNumber(n));
// This code is contributed by suresh07
</script>
Time Complexity: O(1)
Auxiliary Space: O(1)
Using __builtin_clz(x) (GCC builtin function)
Say for a fixed integer (32 bits), count the number of leading zeroes by using the built-in function and subtract it from 31 to get the position of MSB from left, then return the MSB using left shift operation on 1.
An efficient solution for a fixed size integer (say 32 bits) is to one by one set bits, then add 1 so that only the bit after MSB is set. Finally right shift by 1 and return the answer. This solution does not require any condition checking.
C++
// CPP program to find MSB
// number for a given POSITIVE n.
#include <bits/stdc++.h>
using namespace std;
int setBitNumber(int n)
{
// calculate the number
// of leading zeroes
int k = __builtin_clz(n);
// To return the value
// of the number with set
// bit at (31 - k)-th position
// assuming 32 bits are used
return 1 << (31 - k);
}
// Driver code
int main()
{
int n = 273;
cout << setBitNumber(n);
n = ~(int)0; // test for possible overflow
cout << "\n" << (unsigned int)setBitNumber(n);
return 0;
}
import java.lang.*;
public class Main {
public static int setBitNumber(int n) {
// calculate the number
// of leading zeroes
int k = Integer.numberOfLeadingZeros(n);
// To return the value
// of the number with set
// bit at (31 - k)-th position
// assuming 32 bits are used
return 1 << (31 - k);
}
// Driver code
public static void main(String[] args) {
int n = 273;
System.out.println(setBitNumber(n));
n = ~(int)0; // test for possible overflow
System.out.println((int)setBitNumber(n));
}
}
// This code is Contributed by 'Shiv1o43g'
using System;
public class MainClass {
public static int SetBitNumber(int n)
{
// calculate the number
// of leading zeroes
int k = 32 - Convert.ToString(n, 2).Length;
// To return the value
// of the number with set
// bit at (31 - k)-th position
// assuming 32 bits are used
return 1 << (31 - k);
}
// Driver code
public static void Main()
{
int n = 273;
Console.WriteLine(SetBitNumber(n));
n = ~(int)0; // test for possible overflow
Console.WriteLine((uint)SetBitNumber(n));
}
}
// This code is contributed by user_dtewbxkn77n
function setBitNumber(n) {
// calculate the number of leading zeroes
let k = 31 - Math.floor(Math.log2(n));
// To return the value of the number with set
// bit at (31 - k)-th position
return 1 << k;
}
// Driver code
let n = 273;
console.log(setBitNumber(n)); // expected output: 256
n = ~(0); // test for possible overflow
console.log(setBitNumber(n)); // expected output: 2147483648
Time Complexity: O(1)
Auxiliary Space: O(1)
Another Approach: Given a number n. First, find the position of the most significant set bit and then compute the value of the number with a set bit at k-th position.
Thanks Rohit Narayan for suggesting this method.
C++
// CPP program to find MSB
// number for given POSITIVE n.
#include <bits/stdc++.h>
using namespace std;
int setBitNumber(int n)
{
//this will be the answer going to return
//This will work for 64-bit if using with long long
//while in-built functions overflow
int ans = 1;
while (n) {
ans *= 2;
n /= 2;
}
ans/=2;
return ans;
}
// Driver code
int main()
{
int n = 273;
cout << setBitNumber(n);
return 0;
}
#include <stdio.h>
#include <math.h>
int setBitNumber(int n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return (1 << msb);
}
int main() {
int n = 273;
printf("%d",setBitNumber(n));
return 0;
}
// Java program to find MSB
// number for given n.
class GFG {
static int setBitNumber(int n)
{
// To find the position of the
// most significant set bit
int k = (int)(Math.log(n) / Math.log(2));
// To return the value of the number
// with set bit at k-th position
return 1 << k;
}
// Driver code
public static void main(String arg[])
{
int n = 273;
System.out.print(setBitNumber(n));
}
}
# Python program to find
# MSB number for given n.
import math
def setBitNumber(n):
# To find the position of
# the most significant
# set bit
k = int(math.log(n, 2))
# To return the value
# of the number with set
# bit at k-th position
return 1 << k
# Driver code
n = 273
print(setBitNumber(n))
// C# program to find MSB
// number for given n.
using System;
public class GFG {
static int setBitNumber(int n)
{
// To find the position of the
// most significant set bit
int k = (int)(Math.Log(n) / Math.Log(2));
// To return the value of the number
// with set bit at k-th position
return 1 << k;
}
// Driver code
static public void Main()
{
int n = 273;
Console.WriteLine(setBitNumber(n));
}
}
<?php
// PHP program to find MSB
// number for given n.
function setBitNumber($n)
{
// To find the position
// of the most significant
// set bit
$k =(int)(log($n, 2));
// To return the value
// of the number with set
// bit at k-th position
return 1 << $k;
}
// Driver code
$n = 273;
echo setBitNumber($n);
// This code is contributed
// by jit_t.
?>
<script>
// Javascript program to find
// MSB number for given n.
function setBitNumber(n)
{
// To find the position of the
// most significant set bit
let k = parseInt(Math.log(n) / Math.log(2), 10);
// To return the value of the number
// with set bit at k-th position
return 1 << k;
}
let n = 273;
document.write(setBitNumber(n));
</script>
Time Complexity: O(logn)
Auxiliary Space: O(1)
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