[LeetCode Sync] Runtime - 3 ms (64.12%), Memory - 19.7 MB (42.18%)

Author: NinePiece2

Solutions/0136-single-number/README.md

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+<p>Given a <strong>non-empty</strong>&nbsp;array of integers <code>nums</code>, every element appears <em>twice</em> except for one. Find that single one.</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>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [4,1,2,1,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+</div>
+
+<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>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>
+	<li>Each element in the array appears twice except for one element which appears only once.</li>
+</ul>

Solutions/0136-single-number/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)