[LeetCode Sync] Runtime - 0 ms (100.00%), Memory - 19.1 MB (65.46%)

Author: NinePiece2

Solutions/0260-single-number-iii/README.md

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+<p>Given an integer array <code>nums</code>, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in <strong>any order</strong>.</p>
+
+<p>You must write an&nbsp;algorithm that runs in linear runtime complexity and uses&nbsp;only constant extra space.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,1,3,2,5]
+<strong>Output:</strong> [3,5]
+<strong>Explanation: </strong> [5, 3] is also a valid answer.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-1,0]
+<strong>Output:</strong> [-1,0]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,1]
+<strong>Output:</strong> [1,0]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li>Each integer in <code>nums</code> will appear twice, only two integers will appear once.</li>
+</ul>

Solutions/0260-single-number-iii/solution.py

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+class Solution:
+    def singleNumber(self, nums: List[int]) -> List[int]:
+        xor_val = reduce(xor, nums)
+        val1 = 0
+        low_bit = xor_val & -xor_val
+
+        for num in nums:
+            if num & low_bit:
+                val1 ^= num
+        val2 = xor_val ^ val1
+
+        return [val1, val2]