2873. Maximum Value of an Ordered Triplet I

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 <= 100
• 1 <= nums[i] <= 106

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const maximumTripletValue = function (nums) {
  let maxNum = 0;
  let maxDiff = 0;
  let result = 0;

  for (const num of nums) {
    result = Math.max(maxDiff * num, result);
    maxDiff = Math.max(maxNum - num, maxDiff);
    maxNum = Math.max(num, maxNum);
  }

  return result;
};