2873

2873. Maximum Value of an Ordered Triplet I (Easy)

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

Similar Questions:

• Number of Arithmetic Triplets (Easy)

Hints:

• Use three nested loops to find all the triplets.

Solution 1.

// OJ: https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-i
// Author: github.com/lzl124631x
// Time: O(N^3)
// Space: O(1)
class Solution {
public:
    long long maximumTripletValue(vector<int>& A) {
        long long ans = 0, N = A.size();
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                for (int k = j + 1; k < N; ++k) {
                    ans = max(ans, (long long)(A[i] - A[j]) * A[k]);
                }
            }
        }
        return ans;
    }
};

Solution 2. DP

// OJ: https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-i
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-i/solutions/4111950/java-c-python-one-pass-o-n/
class Solution {
public:
    long long maximumTripletValue(vector<int>& A) {
        long long ans = 0, maxa = 0, maxab = 0;
        for (long long n : A) {
            ans = max(ans, (long long)maxab * n);
            maxab = max(maxab, maxa - n);
            maxa = max(maxa, n);
        }
        return ans;
    }
};