[LeetCode Sync] Runtime - 0 ms (100.00%), Memory - 19.7 MB (18.56%)

Author: NinePiece2

Solutions/0137-single-number-ii/README.md

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+<p>Given an integer array <code>nums</code> where&nbsp;every element appears <strong>three times</strong> except for one, which appears <strong>exactly once</strong>. <em>Find the single element and return it</em>.</p>
+
+<p>You must&nbsp;implement a solution with a linear runtime complexity and use&nbsp;only constant&nbsp;extra space.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<pre><strong>Input:</strong> nums = [2,2,3,2]
+<strong>Output:</strong> 3
+</pre><p><strong class="example">Example 2:</strong></p>
+<pre><strong>Input:</strong> nums = [0,1,0,1,0,1,99]
+<strong>Output:</strong> 99
+</pre>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &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 element in <code>nums</code> appears exactly <strong>three times</strong> except for one element which appears <strong>once</strong>.</li>
+</ul>

Solutions/0137-single-number-ii/solution.py

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+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        count = Counter(nums)
+        return next(key for key, value in count.items() if value == 1)