Find The Original Array of Prefix Xor
<p>You are given an <strong>integer</strong> array <code>pref</code> of size <code>n</code>. Find and return <em>the array </em><code>arr</code><em> of size </em><code>n</code><em> that satisfies</em>:</p>

<ul>
	<li><code>pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i]</code>.</li>
</ul>

<p>Note that <code>^</code> denotes the <strong>bitwise-xor</strong> operation.</p>

<p>It can be proven that the answer is <strong>unique</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> pref = [5,2,0,3,1]
<strong>Output:</strong> [5,7,2,3,2]
<strong>Explanation:</strong> From the array [5,7,2,3,2] we have the following:
- pref[0] = 5.
- pref[1] = 5 ^ 7 = 2.
- pref[2] = 5 ^ 7 ^ 2 = 0.
- pref[3] = 5 ^ 7 ^ 2 ^ 3 = 3.
- pref[4] = 5 ^ 7 ^ 2 ^ 3 ^ 2 = 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> pref = [13]
<strong>Output:</strong> [13]
<strong>Explanation:</strong> We have pref[0] = arr[0] = 13.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= pref.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= pref[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Bit Manipulation; Array
function findArray(pref: number[]): number[] {
    let ans = pref.slice();
    for (let i = 1; i < pref.length; i++) {
        ans[i] = pref[i - 1] ^ pref[i];
    }
    return ans;
}
