Check whether kth bit is set or not

Problem Link

Check whether K-th bit is set or not – GFG

Problem Statement

Given a number n and a value k, check whether the k-th bit (0-indexed from the right) in the binary representation of n is set (i.e., is 1) or not.

Return:

• true if the bit is set

• false otherwise

Examples

Example 1:

Input: n = 4, k = 0
Output: false
Explanation: Binary of 4 = 100. 0-th bit (LSB) is 0.

Example 2:

Input: n = 4, k = 2
Output: true
Explanation: Binary of 4 = 100. 2nd bit is 1.

Constraints

• 0 <= n <= 10⁹

• 0 <= k <= 31

Intuition

In binary, each bit represents a power of two. To check if the k-th bit is set, we:

1. Create a bitmask with 1 at the k-th position using left shift: 1 << k.

2. Use bitwise AND with n:

   • If the result is non-zero, the k-th bit is set.

   • Otherwise, it’s not set.

Approach: Bitwise AND with Mask

• Construct a number with only the k-th bit set: 1 << k.

• Perform: n & (1 << k)

  • If result is non-zero → bit is set.

  • Else → not set.

Java Code

class CheckBit {
    static boolean checkKthBit(int n, int k) {
         int check = 1 << k;
         return ((n & check) == check);
    }
}

Time and Space Complexity

Dry Run

Input: n = 4, k = 2

• Binary of n: 100

• Bitmask: 1 << 2 = 100 (binary)

• n & (1 << k) = 100 & 100 = 100 → Non-zero → Bit is set → true

Input: n = 4, k = 0

• Binary of n: 100

• Bitmask: 1 << 0 = 001 (binary)

• n & (1 << 0) = 100 & 001 = 000 → Zero → Bit not set → false

Conclusion

This problem is a fundamental bit manipulation exercise often used to:

• Teach bitmasking

• Check understanding of shift operations and binary logic

It’s a constant-time, zero-extra-space solution and helps form the basis for more complex bit-based algorithms.