<h1 id="heading-problem">Problem</h1>
<p><a target="_blank" href="https://leetcode.com/problems/hamming-distance/">Problem_Link</a></p>
<h1 id="heading-my-attempt">My Attempt</h1>
<h2 id="heading-brainstorming">Brainstorming</h2>
<p>I don't know Hamming distance.
I don't know java bit operation.</p>
<p>what I can think is that &gt;&gt; two numbers and check one by one.</p>
<h3 id="heading-result-code">Result code</h3>
<pre><code class="lang-java">
</code></pre>
<p>fail</p>
<h1 id="heading-study">Study</h1>
<p>Hamming distance : Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.</p>
<p>"karolin" and "kathrin" is 3.</p>
<p>"karolin" and "kerstin" is 3.</p>
<p>"kathrin" and "kerstin" is 4.</p>
<p>0000 and 1111 is 4.</p>
<p>2173896 and 2233796 is 3.</p>
<p>Thus, question want to count two bit numbers not same.</p>
<p>xor gate</p>
<p>0^0=0</p>
<p>0^1=1</p>
<p>1^0=1</p>
<p>1^1=0</p>
<p>xor will make 1's what need to count. and count 1's will be answer.</p>
<h1 id="heading-solution-time-space">Solution (time, space)</h1>
<h2 id="heading-on-on">O(n), O(n)</h2>
<pre><code class="lang-java"><span class="hljs-keyword">public</span> <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span> </span>{
    <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-keyword">int</span> <span class="hljs-title">hammingDistance</span><span class="hljs-params">(<span class="hljs-keyword">int</span> x, <span class="hljs-keyword">int</span> y)</span> </span>{
        <span class="hljs-keyword">return</span> Integer.bitCount(x ^ y);
    }
}
</code></pre>


# Problem

[Problem_Link](https://leetcode.com/problems/hamming-distance/ )

# My Attempt
## Brainstorming
I don't know Hamming distance.
I don't know java bit operation.

what I can think is that >> two numbers and check one by one.



### Result code
```java

```
fail
# Study



Hamming distance : Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.

"karolin" and "kathrin" is 3.

"karolin" and "kerstin" is 3.

"kathrin" and "kerstin" is 4.

0000 and 1111 is 4.

2173896 and 2233796 is 3.

Thus, question want to count two bit numbers not same.

xor gate

0^0=0

0^1=1

1^0=1

1^1=0

xor will make 1's what need to count. and count 1's will be answer.


# Solution (time, space)
## O(n), O(n)
```java
public class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x ^ y);
    }
}
```



461. Hamming Distance

Problem

My Attempt

Brainstorming

Result code

Study

Solution (time, space)

O(n), O(n)