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 (0indexed), 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
 The
rightmostActiveBit
function takes an integern
as input.  It checks if
n
is less than or equal to 0. If true, it returns as there is no rightmost set bit.  The expression
n & n
gives a number with only the rightmost set bit ofn
turned on.  The logarithm base 2 of
n & n
calculates the position of the rightmost set bit.  Adding 1 to the result provides the position of the rightmost set bit.
 The function prints the original number and the calculated result.
Code Solution

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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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