Position of rightmost common bit in two numbers

Given two non-negative numbers m and n. Find the position of rightmost same bit in the binary representation of the numbers.

Examples:  

Input : m = 10, n = 9
Output : 3
(10)10 = (1010)2
(9)10 = (1001)2
It can be seen that the 3rd bit
from the right is same.

Input : m = 16, n = 7
Output : 4
(16)10 = (10000)2
(7)10 = (111)2, can also be written as
 = (00111)2
It can be seen that the 4th bit
from the right is same.

Approach: Get the bitwise xor of m and n. Let it be xor_value = m ^ n. Now, get the position of rightmost unset bit in xor_value.

Explanation: The bitwise xor operation produces a number which has unset bits only at the positions where the bits of m and n are same. Thus, the position of rightmost unset bit in xor_value gives the position of rightmost same bit.

Output: 

Position = 4

Time Complexity: O(1)

Auxiliary Space: O(1)

Alternate Approach: Until both the value becomes zero, check last bits of both numbers and right shift. At any moment, both bits are same, return counter.

Explanation: Rightmost bit of two values m and n are equal only when both values are either odd or even. 

Output: 

Position = 4

Time Complexity: O(1)

Auxiliary Space: O(1)