# Find position of rightmost set bit

The problem of finding the position of the rightmost set bit in a given integer is a common task in computer programming. This position refers to the position of the least significant bit (LSB) that is set to 1 in the binary representation of the number.

## Problem Statement

Given an integer `n`, the task is to find the position of the rightmost set bit in its binary representation.

## Example

Let's consider an example with `n = 320`.

Binary representation of 320: 00101000000

The rightmost set bit is at position 7 (0-indexed), counting from the right. So, the output for this example should be `Number: 320, Result: 7`.

## Idea to Solve The idea to solve this problem is to use the logarithm property. The position of the rightmost set bit in a binary number can be calculated using the formula `log2(n & -n) + 1`.

## Pseudocode

``````function rightmostActiveBit(n):
if n <= 0:
return

result = log2(n & -n) + 1
print("Number:", n, "Result:", result)``````

## Algorithm Explanation

1. The `rightmostActiveBit` function takes an integer `n` as input.
2. It checks if `n` is less than or equal to 0. If true, it returns as there is no rightmost set bit.
3. The expression `n & -n` gives a number with only the rightmost set bit of `n` turned on.
4. The logarithm base 2 of `n & -n` calculates the position of the rightmost set bit.
5. Adding 1 to the result provides the position of the rightmost set bit.
6. The function prints the original number and the calculated result.

## Resultant Output Explanation

The output displays the input number and the calculated position of the rightmost set bit. For example, for `rightmostActiveBit(320)`, the output is `Number: 320, Result: 7`, which means the rightmost set bit in the binary representation of 320 is at position 7.

## Time Complexity

The time complexity of this algorithm is O(1), as it involves basic arithmetic and logarithmic calculations that do not depend on the input size.

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