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authorJack Sangdahl <[email protected]>2025-04-01 21:04:34 -0600
committerJack Sangdahl <[email protected]>2025-04-01 21:04:34 -0600
commit25c3798c3d27de42fa7019951f86756f9b65b238 (patch)
tree3b9f545255aa43e934bce7093f4c00ef6a19b04d /max_value_ordered_triplet1.py
parentb957ec3ee381247dc62dbbffcabcc4badf1235db (diff)
downloadleetcode-25c3798c3d27de42fa7019951f86756f9b65b238.tar.gz
leetcode-25c3798c3d27de42fa7019951f86756f9b65b238.zip
2873. maximum value of an ordered triplet 1
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+# 2873. Maximum Value of an Ordered Triplet I
+
+"""
+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]'
+"""
+
+
+class Solution(object):
+ def maximumTripletValue(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ n = len(nums)
+ if n < 3:
+ return 0
+ ans, curr, diff = 0, nums[0], 0
+ for i in range(1, n):
+ ans = max(ans, diff * nums[i])
+ diff = max(diff, curr - nums[i])
+ curr = max(curr, nums[i])
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maximumTripletValue(nums=[12, 6, 1, 2, 7]))
+ print(obj.maximumTripletValue(nums=[1, 10, 3, 4, 19]))
+ print(obj.maximumTripletValue(nums=[1, 2, 3]))