Find position of the only set bit

Difficulty Level : Easy
Last Updated : 21 Jun, 2021

Given a number N having only one ‘1’ and all other ’0’s in its binary representation, find position of the only set bit. If there are 0 or more than 1 set bit the answer should be -1. Position of set bit ‘1’ should be counted starting with 1 from the LSB side in the binary representation of the number.

Source: Microsoft Interview | 18

Examples:-

Input:
N = 2
Output:
2
Explanation:
2 is represented as "10" in Binary.
As we see there's only one set bit
and it's in Position 2 and thus the
Output 2.

here is another example

Input:
N = 5
Output:
-1
Explanation:
5 is represented as "101" in Binary.
As we see there's two set bits
and thus the Output -1.

The idea is to start from the rightmost bit and one by one check value of every bit. Following is a detailed algorithm.
1) If number is power of two then and then only its binary representation contains only one ‘1’. That’s why check whether the given number is a power of 2 or not. If given number is not a power of 2, then print error message and exit.
2) Initialize two variables; i = 1 (for looping) and pos = 1 (to find position of set bit)
3) Inside loop, do bitwise AND of i and number ‘N’. If value of this operation is true, then “pos” bit is set, so break the loop and return position. Otherwise, increment “pos” by 1 and left shift i by 1 and repeat the procedure.

C++

// C++ program to find position of only set bit in a given number
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to check whether n is a power of 2 or not.
// See http://goo.gl/17Arj
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
// Returns position of the only set bit in 'n'
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
 
    unsigned i = 1, pos = 1;
 
    // Iterate through bits of n till we find a set bit
    // i&n will be non-zero only when 'i' and 'n' have a set bit
    // at same position
    while (!(i & n)) {
        // Unset current bit and set the next bit in 'i'
        i = i << 1;
 
        // increment position
        ++pos;
    }
 
    return pos;
}
 
// Driver program to test above function
int main(void)
{
    int n = 16;
    int pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C

// C program to find position of only set bit in a given number
#include <stdio.h>
 
// A utility function to check whether n is a power of 2 or not.
// See http://goo.gl/17Arj
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
// Returns position of the only set bit in 'n'
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
 
    unsigned i = 1, pos = 1;
 
    // Iterate through bits of n till we find a set bit
    // i&n will be non-zero only when 'i' and 'n' have a set bit
    // at same position
    while (!(i & n)) {
        // Unset current bit and set the next bit in 'i'
        i = i << 1;
 
        // increment position
        ++pos;
    }
 
    return pos;
}
 
// Driver program to test above function
int main(void)
{
    int n = 16;
    int pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    return 0;
}

Java

// Java program to find position of only set bit in a given number
class GFG {
 
    // A utility function to check whether n is a power of 2 or not.
    // See http://goo.gl/17Arj
    static boolean isPowerOfTwo(int n)
    {
        return (n > 0 && ((n & (n - 1)) == 0)) ? true : false;
    }
 
    // Returns position of the only set bit in 'n'
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
 
        int i = 1, pos = 1;
 
        // Iterate through bits of n till we find a set bit
        // i&n will be non-zero only when 'i' and 'n' have a set bit
        // at same position
        while ((i & n) == 0) {
            // Unset current bit and set the next bit in 'i'
            i = i << 1;
 
            // increment position
            ++pos;
        }
 
        return pos;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 16;
        int pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
    }
}
 
// This code is contributed by mits

Python3

# Python3 program to find position of
# only set bit in a given number
 
# A utility function to check
# whether n is power of 2 or
# not.
def isPowerOfTwo(n):
    return (True if(n > 0 and
                   ((n & (n - 1)) > 0))
                 else False);
     
# Returns position of the
# only set bit in 'n'
def findPosition(n):
    if (isPowerOfTwo(n) == True):
        return -1;
 
    i = 1;
    pos = 1;
 
    # Iterate through bits of n
    # till we find a set bit i&n
    # will be non-zero only when
    # 'i' and 'n' have a set bit
    # at same position
    while ((i & n) == 0):
         
        # Unset current bit and
        # set the next bit in 'i'
        i = i << 1;
 
        # increment position
        pos += 1;
 
    return pos;
 
# Driver Code
n = 16;
pos = findPosition(n);
if (pos == -1):
    print("n =", n, ", Invalid number");
else:
    print("n =", n, ", Position ", pos);
 
n = 12;
pos = findPosition(n);
if (pos == -1):
    print("n =", n, ", Invalid number");
else:
    print("n =", n, ", Position ", pos);
 
n = 128;
pos = findPosition(n);
if (pos == -1):
    print("n =", n, ", Invalid number");
else:
    print("n =", n, ", Position ", pos);
 
# This code is contributed by mits

C#

// C# program to find position of only set bit in a given number
using System;
 
class GFG {
 
    // A utility function to check whether n is a power of 2 or not.
    // See http://goo.gl/17Arj
    static bool isPowerOfTwo(int n)
    {
        return (n > 0 && ((n & (n - 1)) == 0)) ? true : false;
    }
 
    // Returns position of the only set bit in 'n'
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
 
        int i = 1, pos = 1;
 
        // Iterate through bits of n till we find a set bit
        // i&n will be non-zero only when 'i' and 'n' have a set bit
        // at same position
        while ((i & n) == 0) {
            // Unset current bit and set the next bit in 'i'
            i = i << 1;
 
            // increment position
            ++pos;
        }
 
        return pos;
    }
 
    // Driver code
    static void Main()
    {
        int n = 16;
        int pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
    }
}
 
// This code is contributed by mits

PHP

<?php
// PHP program to find position of
// only set bit in a given number
 
// A utility function to check
// whether n is power of 2 or
// not. See http://goo.gl/17Arj
function isPowerOfTwo($n)
{
    return $n && (!($n & ($n - 1)));
     
}
 
// Returns position of the
// only set bit in 'n'
function findPosition($n)
{
    if (!isPowerOfTwo($n))
        return -1;
 
    $i = 1;
    $pos = 1;
 
    // Iterate through bits of n
    // till we find a set bit i&n
    // will be non-zero only when
    // 'i' and 'n' have a set bit
    // at same position
    while (!($i & $n))
    {
        // Unset current bit and
        // set the next bit in 'i'
        $i = $i << 1;
 
        // increment position
        ++$pos;
    }
 
    return $pos;
}
 
// Driver Code
$n = 16;
$pos = findPosition($n);
if (($pos == -1) == true)
        echo "n =", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n, ", ",
             " Position ", $pos, "\n";
 
$n = 12;
$pos = findPosition($n);
if(($pos == -1) == true)    
        echo "n = ", $n, ", ",
             " Invalid number", "\n";
else
        echo "n =", $n, ", ",
             " Position ", $pos, "\n";
$n = 128;
$pos = findPosition($n);
if(($pos == -1) == true)    
        echo "n =", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n, ", ",
             " Position ", $pos, "\n";
 
// This code is contributed by ajit
?>

Javascript

<script>
 
// JavaScript program to find position of
// only set bit in a given number
 
// A utility function to check
// whether n is power of 2 or
// not.
function isPowerOfTwo(n){
     
    return (n > 0 && ((n & (n - 1)) == 0)) ? true : false;
}
     
// Returns position of the
// only set bit in 'n'
function findPosition(n){
    if (isPowerOfTwo(n) == false)
        return -1;
 
    var i = 1;
    var pos = 1;
 
    // Iterate through bits of n
    // till we find a set bit i&n
    // will be non-zero only when
    // 'i' and 'n' have a set bit
    // at same position
    while ((i & n) == 0){
         
        // Unset current bit and
        // set the next bit in 'i'
        i = i << 1;
 
        // increment position
        pos += 1;
    }
    return pos;
}
 
// Driver Code
var n = 16;
var pos = findPosition(n);
if (pos == -1)
    document.write("n =" + n + ", Invalid number");
else
    document.write("n =" + n + ", Position " + pos);
    document.write("<br>");
 
n = 12;
pos = findPosition(n);
if (pos == -1)
    document.write("n =" + n + ", Invalid number");
else
    document.write("n =" + n + ", Position ", pos);
    document.write("<br>");
 
n = 128;
pos = findPosition(n);
if (pos == -1)
    document.write("n =" + n + ", Invalid number");
else
    document.write("n =" + n + ", Position " + pos);
 
// This code is contributed by AnkThon
 
</script>

Output :

n = 16, Position 5
n = 12, Invalid number
n = 128, Position 8

Following is another method for this problem. The idea is to one by one right shift the set bit of given number ‘n’ until ‘n’ becomes 0. Count how many times we shifted to make ‘n’ zero. The final count is the position of the set bit.

C++

// C++ program to find position of only set bit in a given number
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to check whether n is power of 2 or not
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
// Returns position of the only set bit in 'n'
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
 
    unsigned count = 0;
 
    // One by one move the only set bit to right till it reaches end
    while (n)
    {
        n = n >> 1;
 
        // increment count of shifts
        ++count;
    }
 
    return count;
}
 
// Driver code
int main(void)
{
    int n = 0;
    int pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
                    cout<<"n = "<<n<<", Position "<< pos<<endl;
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
               cout<<"n = "<<n<<", Position "<< pos<<endl;
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
                cout<<"n = "<<n<<", Position "<< pos<<endl;
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C

// C program to find position of only set bit in a given number
#include <stdio.h>
 
// A utility function to check whether n is power of 2 or not
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
// Returns position of the only set bit in 'n'
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
 
    unsigned count = 0;
 
    // One by one move the only set bit to right till it reaches end
    while (n) {
        n = n >> 1;
 
        // increment count of shifts
        ++count;
    }
 
    return count;
}
 
// Driver program to test above function
int main(void)
{
    int n = 0;
    int pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    return 0;
}

Java

// Java program to find position of only
// set bit in a given number
 
class GFG {
 
    // A utility function to check whether
    // n is power of 2 or not
    static boolean isPowerOfTwo(int n)
    {
        return n > 0 && ((n & (n - 1)) == 0);
    }
 
    // Returns position of the only set bit in 'n'
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
 
        int count = 0;
 
        // One by one move the only set bit
        // to right till it reaches end
        while (n > 0) {
            n = n >> 1;
 
            // increment count of shifts
            ++count;
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 0;
        int pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number");
        else
            System.out.println("n = " + n + ", Position " + pos);
    }
}
 
// This code is contributed by mits

Python3

# Python 3 program to find position
# of only set bit in a given number
 
# A utility function to check whether
# n is power of 2 or not
def isPowerOfTwo(n) :
 
    return (n and ( not (n & (n-1))))
 
# Returns position of the only set bit in 'n'
def findPosition(n) :
 
    if not isPowerOfTwo(n) :
        return -1
 
    count = 0
 
    # One by one move the only set bit to
    # right till it reaches end
    while (n) :
         
        n = n >> 1
 
        # increment count of shifts
        count += 1
 
    return count
 
 
# Driver program to test above function
if __name__ == "__main__" :
    n = 0
    pos = findPosition(n)
 
    if pos == -1 :
        print("n =", n, "Invalid number")
    else :
        print("n =", n, "Position", pos)
 
    n = 12
    pos = findPosition(n)
 
    if pos == -1 :
        print("n =", n, "Invalid number")
    else :
        print("n =", n, "Position", pos)
 
    n = 128
    pos = findPosition(n)
 
    if pos == -1 :
        print("n =", n, "Invalid number")
    else :
        print("n =", n, "Position", pos)
    
# This code is contributed by ANKITRAI1

C#

// C# program to find position of only
// set bit in a given number
using System;
 
class GFG {
 
    // A utility function to check whether
    // n is power of 2 or not
    static bool isPowerOfTwo(int n)
    {
        return n > 0 && ((n & (n - 1)) == 0);
    }
 
    // Returns position of the only set bit in 'n'
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
 
        int count = 0;
 
        // One by one move the only set bit
        // to right till it reaches end
        while (n > 0) {
            n = n >> 1;
 
            // increment count of shifts
            ++count;
        }
 
        return count;
    }
 
    // Driver code
    static void Main()
    {
        int n = 0;
        int pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
    }
}
 
// This code is contributed by mits

PHP

<?php
// PHP program to find position of
// only set bit in a given number
 
// A utility function to check
// whether n is power of 2 or not
function isPowerOfTwo($n)
{
    return $n && (! ($n & ($n - 1)));
}
 
// Returns position of the
// only set bit in 'n'
function findPosition($n)
{
    if (!isPowerOfTwo($n))
        return -1;
 
    $count = 0;
 
    // One by one move the only set
    // bit to right till it reaches end
    while ($n)
    {
        $n = $n >> 1;
 
        // increment count of shifts
        ++$count;
    }
 
    return $count;
}
 
// Driver Code
$n = 0;
$pos = findPosition($n);
if(($pos == -1) == true)
    echo "n = ", $n, ", ",
         " Invalid number", "\n";
else
    echo "n = ", $n, ", ",
         " Position ", $pos, "\n";
 
$n = 12;
$pos = findPosition($n);
if (($pos == -1) == true)
        echo "n = ", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n,
             " Position ", $pos, "\n";
 
$n = 128;
$pos = findPosition($n);
    if(($pos == -1) == true)
        echo "n = ", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n, ", ",
             " Position ", $pos, "\n";
         
// This code is contributed by ajit
?>

Javascript

<script>
 
// JavaScript program to find position
// of only set bit in a given number
 
// A utility function to check whether
// n is power of 2 or not
function isPowerOfTwo(n) {
 
    return (n && ( !(n & (n-1))))
}
 
// Returns position of the only set bit in 'n'
function findPosition(n) {
 
    if (!isPowerOfTwo(n))
        return -1
 
    var count = 0
 
    // One by one move the only set bit to
    // right till it reaches end
    while (n) {
         
        n = n >> 1
 
        // increment count of shifts
        count += 1
    }
 
    return count
 
}
 
// Driver program to test above function
var n = 0
var pos = findPosition(n)
 
if (pos == -1)
    document.write("n = ", n, ", Invalid number ")
else
    document.write("n =", n, ", Position ", pos)
     
document.write("<br>")
 
n = 12
pos = findPosition(n)
 
if (pos == -1)
    document.write("n = ", n, ", Invalid number")
else
    document.write("n = ", n, ", Position ", pos)
 
document.write("<br>")
 
n = 128
pos = findPosition(n)
 
if (pos == -1)
    document.write("n = ", n, ", Invalid number")
else
    document.write("n = ", n, ", Position ", pos)
   
document.write("<br>")
 
// This code is contributed by AnkThon
 
</script>

Output :

n = 0, Invalid number
n = 12, Invalid number
n = 128, Position 8

We can also use log base 2 to find the position. Thanks to Arunkumar for suggesting this solution.

C++

#include <bits/stdc++.h>
using namespace std;
 
unsigned int Log2n(unsigned int n)
{
    return (n > 1) ? 1 + Log2n(n / 2) : 0;
}
 
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
    return Log2n(n) + 1;
}
 
// Driver code
int main(void)
{
    int n = 0;
    int pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
            cout<<"n = "<<n<<", Position "<<pos<<" \n";
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
            cout<<"n = "<<n<<", Position "<<pos<<" \n";
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? cout<<"n = "<<n<<", Invalid number\n" :
            cout<<"n = "<<n<<", Position "<<pos<<" \n";
 
    return 0;
}
 
 
// This code is contributed by rathbhupendra

C

#include <stdio.h>
 
unsigned int Log2n(unsigned int n)
{
    return (n > 1) ? 1 + Log2n(n / 2) : 0;
}
 
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
 
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
    return Log2n(n) + 1;
}
 
// Driver program to test above function
int main(void)
{
    int n = 0;
    int pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? printf("n = %d, Invalid number\n", n) : printf("n = %d, Position %d \n", n, pos);
 
    return 0;
}

Java

// Java program to find position
// of only set bit in a given number
 
class GFG {
    static int Log2n(int n)
    {
        return (n > 1) ? 1 + Log2n(n / 2) : 0;
    }
 
    static boolean isPowerOfTwo(int n)
    {
        return n > 0 && ((n & (n - 1)) == 0);
    }
 
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
        return Log2n(n) + 1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 0;
        int pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number ");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number ");
        else
            System.out.println("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            System.out.println("n = " + n + ", Invalid number ");
        else
            System.out.println("n = " + n + ", Position " + pos);
    }
}
 
// This code is contributed by mits

Python3

# Python program to find position
# of only set bit in a given number
 
def Log2n(n):
    if (n > 1):
        return (1 + Log2n(n / 2))
    else:
        return 0
     
# A utility function to check
# whether n is power of 2 or not   
def isPowerOfTwo(n):
    return n and (not (n & (n-1)) )
     
def findPosition(n):
    if (not isPowerOfTwo(n)):
        return -1
    return Log2n(n) + 1
     
# Driver program to test above function
 
n = 0
pos = findPosition(n)
if(pos == -1):
    print("n =", n, ", Invalid number")
else:
    print("n = ", n, ", Position ", pos)
  
n = 12
pos = findPosition(n)
if(pos == -1):
    print("n =", n, ", Invalid number")
else:
    print("n = ", n, ", Position ", pos)
n = 128
pos = findPosition(n)
if(pos == -1):
    print("n = ", n, ", Invalid number")
else:
    print("n = ", n, ", Position ", pos)
  
# This code is contributed
# by Sumit Sudhakar

C#

// C# program to find position
// of only set bit in a given number
 
using System;
 
class GFG {
    static int Log2n(int n)
    {
        return (n > 1) ? 1 + Log2n(n / 2) : 0;
    }
 
    static bool isPowerOfTwo(int n)
    {
        return n > 0 && ((n & (n - 1)) == 0);
    }
 
    static int findPosition(int n)
    {
        if (!isPowerOfTwo(n))
            return -1;
        return Log2n(n) + 1;
    }
 
    // Driver program to test above function
    static void Main()
    {
        int n = 0;
        int pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number ");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 12;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number ");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
 
        n = 128;
        pos = findPosition(n);
        if (pos == -1)
            Console.WriteLine("n = " + n + ", Invalid number ");
        else
            Console.WriteLine("n = " + n + ", Position " + pos);
    }
}
// This code is contributed by mits

PHP

<?php
// PHP program to find position
// of only set bit in a given number
function Log2n($n)
{
return ($n > 1) ? 1 +
  Log2n($n / 2) : 0;
}
 
function isPowerOfTwo($n)
{
    return $n && (! ($n &
                    ($n - 1)));
}
 
function findPosition($n)
{
    if (!isPowerOfTwo($n))
        return -1;
    return Log2n($n) + 1;
}
 
// Driver Code
$n = 0;
$pos = findPosition($n);
if(($pos == -1) == true)
        echo "n =", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n, ", ",
             " Position n", $pos, "\n";
 
$n = 12;
$pos = findPosition($n);
if(($pos == -1) == true)
            echo "n = ", $n, ", ",
                 " Invalid number", "\n";
        else
            echo "n =", $n, ", ",
                 " Position", $pos, "\n";
 
// Driver Code
$n = 128;
$pos = findPosition($n);
if(($pos == -1) == true)
        echo "n = ", $n, ", ",
             " Invalid number", "\n";
else
        echo "n = ", $n, ", ",
             " Position ", $pos, "\n";
         
// This code is contributed by aj_36
?>

Javascript

<script>
 
// JavaScript program to find position
// of only set bit in a given number
 
function Log2n(n){
    if (n > 1)
        return (1 + Log2n(n / 2))
    else
        return 0
}
     
// A utility function to check
// whether n is power of 2 or not   
function isPowerOfTwo(n){
    return n && ( ! (n & (n-1)) )
}
     
function findPosition(n){
    if (isPowerOfTwo(n) == false)
        return -1
    return Log2n(n) + 1
 }
     
// Driver program to test above function
 
var n = 0
var pos = findPosition(n)
if(pos == -1)
    document.write("n = ", n, ", Invalid number")
else
    document.write("n = ", n, ", Position ", pos)
 
document.write("<br>")
 
n = 12
pos = findPosition(n)
if(pos == -1)
    document.write("n = ", n, ", Invalid number")
else
    document.write("n = ", n, ", Position ", pos)
  
document.write("<br>")
n = 128
pos = findPosition(n)
if(pos == -1)
    document.write("n = ", n, ", Invalid number")
else
    document.write("n = ", n, ", Position ", pos)
  
// This code is contributed AnkThon
 
 
</script>

Output :

n = 0, Invalid number
n = 12, Invalid number
n = 128, Position 8

This article is compiled by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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