3152. Maximum Value Of An Ordered Triplet Ii¶

Difficulty: Medium

3152. Maximum Value of an Ordered Triplet II

Medium

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 <= 10⁵
1 <= nums[i] <= 10⁶

Solution¶

class Solution {
    public long maximumTripletValue(int[] nums) {
        int n = nums.length;
        int pref_maxi[] = new int[n];
        int suff_maxi[] = new int[n];
        int maxi = nums[0];
        for (int i = 0; i < n; i++) {
            maxi = Math.max(maxi, nums[i]);
            pref_maxi[i] = maxi;
        }
        maxi = nums[n - 1];
        for (int i = n - 1; i >= 0; i--) {
            maxi = Math.max(maxi, nums[i]);
            suff_maxi[i] = maxi;
        }
        long res = 0;
        for (int i = 0; i < n; i++) {
            if (i + 2 < n) {
                int left = pref_maxi[i], middle = nums[i + 1], right = suff_maxi[i + 2];
                res = Math.max(res, (left - middle) * 1L * right);
            }
        }
        return res;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here