Add solution for increasing array triplet

Author: kwnna15

array/increasing-triplet/README.md

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+# 334. Increasing Triplet Subsequence
+
+## Description
+See https://leetcode.com/problems/increasing-triplet-subsequence/description/
+
+## Problem
+Given an integer array `nums`, return `true` if there exists a triple of indices `(i, j, k)` such that `i < j < k` and `nums[i] < nums[j] < nums[k]`. If no such indices exists, return `false`.
+
+## Example 1
+
+```
+Input: nums = [1,2,3,4,5]
+Output: true
+Explanation: Any triplet where i < j < k is valid.
+```
+
+## Example 2
+
+```
+Input: nums = [5,4,3,2,1]
+Output: false
+Explanation: No triplet exists.
+```
+
+## Example 3
+
+```
+Input: nums = [2,1,5,0,4,6]
+Output: true
+Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
+```
+
+## Constraints
+
+```
+1 <= nums.length <= 5 * 105
+-231 <= nums[i] <= 231 - 1
+```
+
+**Follow up:** Could you implement a solution that runs in `O(n)` time complexity and `O(1)` space complexity?

array/increasing-triplet/increasingtriplet.py

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+def increasing_triplet(nums: list[int]) -> bool:
+    first = second = float("inf")
+    for n in nums:
+        if n <= first:
+            first = n
+        elif n <= second:
+            second = n
+        else:
+            return True
+    return False

array/increasing-triplet/increasingtriplet_test.py

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+from increasingtriplet import increasing_triplet
+
+
+def test_example_1():
+    nums = [1, 2, 3, 4, 5]
+    assert increasing_triplet(nums) == True
+
+
+def test_example_2():
+    nums = [5, 4, 3, 2, 1]
+    assert increasing_triplet(nums) == False
+
+
+def test_example_3():
+    nums = [2, 1, 5, 0, 4, 6]
+    assert increasing_triplet(nums) == True
+
+
+def test_example_4():
+    nums = [20, 100, 10, 12, 5, 13]
+    assert increasing_triplet(nums) == True
+
+
+def test_example_5():
+    nums = [2, 2, 3, 3, 2, 2, 2, 3, 1]
+    assert increasing_triplet(nums) == False