find-the-xor-of-numbers-which-appear-twice

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Algorithms

You are given an array <code>nums</code>, where each number in the array appears either once or twice.

Return the bitwise <code>XOR</code> of all the numbers that appear twice in the array, or 0 if no number appears twice.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [1,2,1,3]

Output: 1

Explanation:

The only number that appears twice in&nbsp;<code>nums</code>&nbsp;is 1.
</div>

Example 2:

<div class="example-block">
Input: nums = [1,2,3]

Output: 0

Explanation:

No number appears twice in&nbsp;<code>nums</code>.
</div>

Example 3:

<div class="example-block">
Input: nums = [1,2,2,1]

Output: 3

Explanation:

Numbers 1 and 2 appeared twice. <code>1 XOR 2 == 3</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
	<li>Each number in <code>nums</code> appears either once or twice.</li>
</ul>