Description
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 <= 1001 <= nums[i] <= 106
Solutions
Solution 1: Maintaining Prefix Maximum and Maximum Difference
We use two variables mx and mxDiff to maintain the prefix maximum value and maximum difference, respectively, and use a variable ans to maintain the answer. Initially, these variables are all 0.
Next, we iterate through each element x in the array as nums[k]. First, we update the answer ans = max(ans, mxDiff × x). Then we update the maximum difference mxDiff = max(mxDiff, mx - x). Finally, we update the prefix maximum value mx = max(mx, x).
After iterating through all elements, we return the answer ans.
The time complexity is O(n), where n is the length of the array. The space complexity is O(1).
class Solution: def maximumTripletValue(self, nums: List[int]) -> int: ans = mx = mx_diff = 0 for x in nums: ans = max(ans, mx_diff * x) mx_diff = max(mx_diff, mx - x) mx = max(mx, x) return ans(code-box)
