Check If Kth Bit Is Set/Unset

Check if Kth Bit Is Set/Unset

Hamming Distance - Java Solution

#             
## Overview
We try to find a bit in the binary representation of a number at the $k^{th}$ position and check if it is set/unset.




## Problem statement
Given two positive integers (`n` and `k`) we check whether the bit at position `k`, from the right in the binary representation of n, is set `1` or unset `0`.

**Example 01:**

```
Input: n = 5, k = 1 
 
Output: true
```

**Example 02:**

```
Input: n = 10, k = 2
 
Output: true
```

**Example 03:**

```
Input: n = 10, k = 1 
 
Output: false
```



if(n == 0) return;
k = 1
while(true) {
 if((n & (1 << (k - 1))) == 0) 
    increment k
  else {
    return k
  }
}
    

class CheckKthBitSetOrNot {
  private static boolean checkKthBitSet(int n, int k) {
    return (n & (1 << (k - 1))) != 0;
  }

  public static void main(String[] args) {
    System.out.println("n = 5, k = 3 : " + checkKthBitSet(5, 3));
    System.out.println("------------");
    System.out.println("n = 10, k = 2 : " + checkKthBitSet(10, 2));
    System.out.println("------------");
    System.out.println("n = 10, k = 1 : " + checkKthBitSet(10, 1));
  }
}


Input n = 5, k = 3

   n    = 5 = 00000000 00000000 00000000 00000101
   1    = 1 = 00000000 00000000 00000000 00000001
   k    = 3 = 00000000 00000000 00000000 00000011
(k - 1) = 2 = 00000000 00000000 00000000 00000010

(1 << (k - 1)) = 4 = 00000000 00000000 00000000 00000100

         n         = 5 = 00000000 00000000 00000000 00000101
  (1 << (k - 1))   = 4 = 00000000 00000000 00000000 00000100
-------------------------------------------------------------
n & (1 << (k - 1)) = 4 = 00000000 00000000 00000000 00000100
-------------------------------------------------------------

How to check if the kth bit is set/unset

Given integers n and k, determine if the k-th bit from the right in n's binary form is set (1) or unset (0).

How to check if the kth bit is set/unset

Overview

Problem statement

Code

Complexity analysis

Optimal solution (refactored)

Explanation

Case 01

Complexity analysis