@Solution: Maximum Value of an Ordered Triplet II

Author: i-redbyte

Diff for: medium/maximum value of an ordered triplet II/README.md

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+# Maximum Value of an Ordered Triplet II
+
+You are given a 0-indexed integer array nums.
+
+Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a
+negative value, return 0.
+
+The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].
+
+Example 1:
+
+```
+Input: nums = [12,6,1,2,7]
+Output: 77
+Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77. 
+```
+
+Example 2:
+
+```
+Input: nums = [1,10,3,4,19]
+Output: 133
+Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
+It can be shown that there are no ordered triplets of indices with a value greater than 133.
+```
+
+Example 3:
+
+```
+Input: nums = [1,2,3]
+Output: 0
+Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
+```
+
+**Constraints:**
+
+- 3 <= nums.length <= 10^5
+- 1 <= nums[i] <= 10^6
+
+[Link](https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-ii/description)

Diff for: medium/maximum value of an ordered triplet II/solution.py

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+from typing import List
+
+
+class Solution:
+    def maximumTripletValue(self, nums: List[int]) -> int:
+        n = len(nums)
+        result, imax, tmp = 0, 0, 0
+        for i in range(n):
+            result = max(result, tmp * nums[i])
+            tmp = max(tmp, imax - nums[i])
+            imax = max(imax, nums[i])
+        return result
+
+
+s = Solution()
+print(s.maximumTripletValue([12, 6, 1, 2, 7]))
+print(s.maximumTripletValue([1, 10, 3, 4, 19]))
+print(s.maximumTripletValue([1, 2, 3]))
+print(s.maximumTripletValue([1, 3]))