Find most significant set bit of a number

27 May, 2022
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
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++
C
Java
Python3
C#
PHP
Javascript
// 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 codeint 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 coden = 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 codelet n = 0;document.write (setBitNumber(n));     // This code is contributed by Bobby </script>
0
 
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++
C
Java
Python3
C#
PHP
Javascript
// 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 codeint 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 codelet n = 273;document.write(setBitNumber(n)); // This code is contributed by suresh07 </script>
256
 
Time complexity is 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++
C
Java
Python3
C#
PHP
Javascript
// CPP program to find MSB// number for given POSITIVE n.#include <bits/stdc++.h>using namespace std; int setBitNumber(int n){     // To find the position    // of the most significant    // set bit    int k = (int)(log2(n));     // To return the value    // of the number with set    // bit at k-th position    return 1 << k;} // Driver codeint 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 <math.h> int setBitNumber(int n){     // To find the position    // of the most significant    // set bit    int k = (int)(log2(n));     // To return the value    // of the number with set    // bit at k-th position    return 1 << k;}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 coden = 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>
256
 
Another Approach : 
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