[LeetCode Sync] Runtime - 13 ms (79.11%), Memory - 37.4 MB (89.98%)

Author: NinePiece2

Solutions/0334-increasing-triplet-subsequence/README.md

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+<p>Given an integer array <code>nums</code>, return <code>true</code><em> if there exists a triple of indices </em><code>(i, j, k)</code><em> such that </em><code>i &lt; j &lt; k</code><em> and </em><code>nums[i] &lt; nums[j] &lt; nums[k]</code>. If no such indices exists, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> Any triplet where i &lt; j &lt; k is valid.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5,4,3,2,1]
+<strong>Output:</strong> false
+<strong>Explanation:</strong> No triplet exists.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,1,5,0,4,6]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> One of the valid triplet is (3, 4, 5), because nums[3] == 0 &lt; nums[4] == 4 &lt; nums[5] == 6.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Could you implement a solution that runs in <code>O(n)</code> time complexity and <code>O(1)</code> space complexity?

Solutions/0334-increasing-triplet-subsequence/solution.py

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+class Solution:
+    def increasingTriplet(self, nums: List[int]) -> bool:
+        first, second = float('inf'), float('inf')
+        for num in nums:
+            if num <= first:
+                first = num
+            elif num <= second:
+                second = num
+            else:
+                return True
+        return False