Check if n is divisible by power of 2 without using arithmetic operators

Given two positive integers n and m, check whether n is divisible by 2^m without using arithmetic operators

Examples: 

Input: n = 8, m = 2
Output: Yes
Explanation: 2^2=4, and 8 is divisible by 4.

Input: n = 14, m = 3
Output: No
Explanation: 2^3=8, and 14 is not divisible by 8.

Using Bitwise Operators - O(1) Time and O(1) Space

The idea of this approach is based on the bitwise properties of powers of 2. A number is divisible by 2 if its least significant bit (LSB) is 0, divisible by 4 if the two least significant bits are 0, divisible by 8 if the three least significant bits are 0, and so on. Using this property, a number n is divisible by 2^m if its last m bits are all 0.

We can check this using the expression: n & ((1 << m) - 1)

Here’s how it works:

• 1 << m shifts the number 1 to the left by m positions, giving 2^m.
• (1 << m) - 1 creates a mask with the last m bits set to 1.
• n & ((1 << m) - 1) extracts the last m bits of n.
• If the result is 0, then n is divisible by 2^m; otherwise, it is not.

Yes