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Diffstat (limited to 'max_value_ordered_triplet2.py')
-rw-r--r-- | max_value_ordered_triplet2.py | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/max_value_ordered_triplet2.py b/max_value_ordered_triplet2.py new file mode 100644 index 0000000..5673d61 --- /dev/null +++ b/max_value_ordered_triplet2.py @@ -0,0 +1,32 @@ +# 2874. 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]' +""" + + +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])) |