2874

2874. 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

Companies: Media.net

Related Topics:
Array

Similar Questions:

• Trapping Rain Water (Hard)
• Sum of Beauty in the Array (Medium)

Hints:

• Preprocess the prefix maximum array, prefix_max[i] = max(nums[0], nums[1], …, nums[i]) and the suffix maximum array, suffix_max[i] = max(nums[i], nums[i + 1], …, nums[i - 1]).
• For each index j, find two indices i and k such that i < j < k and (nums[i] - nums[j]) * nums[k] is the maximum, using the prefix and suffix maximum arrays.
• For index j, the maximum triplet value is (prefix_max[j - 1] - nums[j]) * suffix_max[j + 1].

Solution 1. Kadane

// OJ: https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-ii
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    long long maximumTripletValue(vector<int>& A) {
        long long ans = 0, mx = A[0], first = A[0] - A[1];
        for (int i = 2, N = A.size(); i < N; ++i) {
            mx = max(mx, (long long)A[i - 2]);
            first = max(first, mx - A[i - 1]);
            ans = max(ans, first * A[i]);
        }
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-ii
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    long long maximumTripletValue(vector<int>& A) {
        long long ans = 0, mx = 0, first = 0;
        for (int n : A) {
            ans = max(ans, first * n);
            first = max(first, mx - n);
            mx = max(mx, (long long)n);
        }
        return ans;
    }
};