Position of the only set bit (original) (raw)

Last Updated : 22 Apr, 2026

Given a number **n containing only 1 set bit in its binary representation, the task is to find the **position of the only set bit. If there are 0 or more than 1 set bits, then return -1.

**Note: Position of set bit '1' should be counted starting with 1 from the LSB side in the binary representation of the number.

**Examples:-

**Input: n = 2
**Output: 2
**Explanation: Binary representation of 2 is 10. We can observe that the only set bit is at position 2 from LSB.

**Input: n = 5
**Output: -1
**Explanation: Binary representation of 5 is 101. There are 2 set bits, so return -1.

Try It Yourself

**Condition for numbers having only 1 set bit:

Numbers having only one set bit will be a power of 2 number ( example, 20 = 1, 21 = 10, 22 = 100, 23 = 1000).

When you subtract 1 from such a number, all bits after the set bit (including the set bit itself) flip (e.g., 4 = 100, 3 = 011). Performing a bitwise AND between n and n-1 results in 0 if n is a power of 2, as the single set bit cancels out.

Table of Content

• Using Left Shift Operator - O(log(n)) time and O(1) space
• Using Right Shift Operator - O(log(n)) time and O(1) space
• Using Log Operator - O(log n) time and O(1) space

Using Left Shift Operator - O(log(n)) time and O(1) space

The idea is to use a loop where we left shift the number 1 and perform a bitwise AND operation with n. If the result is non-zero, the position of the set bit is determined by the number of shifts performed.

Dry run for n = 2 (binary = 10) :

• Check (n == 0 => false) OR ((n & (n - 1)) != 0)
  (n & (n - 1)) = (2 & 1) = 0 => (0 != 0) = false
  So (false OR false = false), hence continue.
• Initialize pos = 1, val = 1 (binary: 01)
• (val & n) = (1 & 2) = 0 so condition true, enter loop
  val = val << 1 = 1 << 1 = 2 (binary: 10), pos = 2
• (val & n) = (2 & 2) = 2 so condition false, exit loop

Final answer is: pos = 2

C++ `

#include <bits/stdc++.h> using namespace std;

int findPosition(int n) { // Check if n has exactly one set bit if (n == 0 || (n & (n - 1)) != 0) return -1;

int pos = 1;
int val = 1;
while ((val & n) == 0)
{
    // shifting left
    val = val << 1;
    pos++;
}
return pos;

}

// Main function int main() { int n = 2; cout << findPosition(n); return 0; }

Java

class GfG {

static int findPosition(int n)
{
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0)
        return -1;

    int pos = 1;
    int val = 1;
    while ((val & n) == 0) {
        // Shifting left
        val = val << 1;
        pos++;
    }
    return pos;
}

// Main function
public static void main(String[] args)
{
    int n = 2;
    System.out.println(findPosition(n));
}

}

Python

def findPosition(n):

# Check if n has exactly one set bit
if n == 0 or (n & (n - 1)) != 0:
    return -1

pos = 1
val = 1
while (val & n) == 0:
    # Shifting left
    val = val << 1
    pos += 1
return pos

Driver Code

if name == "main": n = 2 print(findPosition(n))

C#

using System;

class GfG {

static int findPosition(int n) {
    
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0) return -1; 
    
    int pos = 1;
    int val = 1;
    while ((val & n) == 0) {
        // shifting left
        val = val << 1;
        pos++;
    }
    return pos;
}

// Main function
static void Main() {
    int n = 2;
    Console.WriteLine(findPosition(n));
}

}

JavaScript

function findPosition(n) {

// Check if n has exactly one set bit
if (n === 0 || (n & (n - 1)) !== 0) return -1; 

let pos = 1;
let val = 1;
while ((val & n) === 0) {
    // Shifting left
    val = val << 1;
    pos++;
}
return pos;

}

// Driver Code let n = 2; console.log(findPosition(n));

`

Using Right Shift Operator - O(log(n)) time and O(1) space

The idea is to right shift the number n until the rightmost bit becomes 1. The number of shifts required to reach this point gives the position of the set bit.

Dry run for n = 2 (binary = 10) :

• Check (n == 0 => false) OR ((n & (n - 1)) != 0)
  (n & (n - 1)) = (2 & 1) = 0 => (0 != 0) = false
  So (false OR false = false), hence continue.
• Initialize pos = 1
• (n & 1) = (2 & 1) = 0 so condition is true, enter loop
  n = n >> 1 = 2 >> 1 = 1 (binary: 01), pos = 2
• (n & 1) = (1 & 1) = 1 so condition is false, exit loop

Final answer is: pos = 2

C++ `

#include <bits/stdc++.h> using namespace std;

int findPosition(int n) { // Check if n has exactly one set bit if (n == 0 || (n & (n - 1)) != 0) return -1;

int pos = 1;
while ((n & 1) == 0)
{
    // Shifting right
    n = n >> 1;
    pos++;
}
return pos;

}

// Main function int main() { int n = 2; cout << findPosition(n); return 0; }

Java

class GfG {

static int findPosition(int n)
{
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0)
        return -1;

    int pos = 1;
    while ((n & 1) == 0) {
        
        // Shifting right
        n = n >> 1;
        pos++;
    }
    return pos;
}

// Main function
public static void main(String[] args)
{
    int n = 2;
    System.out.println(findPosition(n));
}

}

Python

def findPosition(n):

# Check if n has exactly one set bit
if n == 0 or (n & (n - 1)) != 0:
    return -1

pos = 1
while (n & 1) == 0:

    # Shifting right
    n = n >> 1
    pos += 1
return pos

Driver code

if name == "main": n = 2 print(findPosition(n))

C#

using System;

class GfG {

static int findPosition(int n)
{
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0)
        return -1;

    int pos = 1;
    while ((n & 1) == 0) {

        // Shifting right
        n = n >> 1;
        pos++;
    }
    return pos;
}

// Main function
static void Main()
{
    int n = 2;
    Console.WriteLine(findPosition(n));
}

}

JavaScript

function findPosition(n) { // Check if n has exactly one set bit if (n === 0 || (n & (n - 1)) !== 0) return -1;

let pos = 1;
while ((n & 1) === 0) {

    // Shifting right
    n = n >> 1;
    pos++;
}
return pos;

}

// Driver code let n = 2; console.log(findPosition(n));

`

Using Log Operator - O(log n) time and O(1) space

The idea is to use the mathematical property that the position of the only set bit in a number n (which is a power of 2) can be found by taking the base-2 logarithm of n and adding 1 (since the position is 1-based).

C++ `

#include <bits/stdc++.h> using namespace std;

int findPosition(int n) { // Check if n has exactly one set bit if (n == 0 || (n & (n - 1)) != 0) return -1;

// Find position of the set bit using log2
return log2(n) + 1;

}

// Main function int main() { int n = 2; cout << findPosition(n); return 0; }

Java

class GfG {

static int findPosition(int n)
{
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0)
        return -1;

    // Find position of the set bit using log2
    return (int)(Math.log(n) / Math.log(2)) + 1;
}

// Main function
public static void main(String[] args)
{
    int n = 2;
    System.out.println(findPosition(n));
}

}

Python

import math

def findPosition(n):

# Check if n has exactly one set bit
if n == 0 or (n & (n - 1)) != 0:
    return -1

# Find position of the set bit using log2
return int(math.log2(n)) + 1

Driver code

if name == "main": n = 2 print(int(findPosition(n)))

C#

using System;

class GfG {

static int findPosition(int n)
{
    // Check if n has exactly one set bit
    if (n == 0 || (n & (n - 1)) != 0)
        return -1;

    // Find position of the set bit using log2
    return (int)(Math.Log(n) / Math.Log(2)) + 1;
}

// Main function
static void Main()
{
    int n = 2;
    Console.WriteLine(findPosition(n));
}

}

JavaScript

function findPosition(n) {

// Check if n has exactly one set bit
if (n === 0 || (n & (n - 1)) !== 0) return -1; 

// Find position of the set bit using log2
return Math.log2(n) + 1;

}

// Driver code let n = 2; console.log(Math.floor(findPosition(n)));

`

**Related Article:

• Program to find whether a given number is power of 2