2874. Maximum Value of an Ordered Triplet II 

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 <= 10⁵
1 <= nums[i] <= 10⁶

Solutions

Solution 1: Maintaining Prefix Maximum and Maximum Difference

We use two variables $\textit{mx}$ and $\textit{mxDiff}$ to maintain the prefix maximum value and maximum difference, respectively, and use a variable $\textit{ans}$ to maintain the answer. Initially, these variables are all $0$.

Next, we iterate through each element $x$ in the array as $\textit{nums}[k]$. First, we update the answer $\textit{ans} = \max(\textit{ans}, \textit{mxDiff} \times x)$. Then we update the maximum difference $\textit{mxDiff} = \max(\textit{mxDiff}, \textit{mx} - x)$. Finally, we update the prefix maximum value $\textit{mx} = \max(\textit{mx}, x)$.

After iterating through all elements, we return the answer $\textit{ans}$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public long maximumTripletValue(int[] nums) {
        long ans = 0, mxDiff = 0;
        int mx = 0;
        for (int x : nums) {
            ans = Math.max(ans, mxDiff * x);
            mxDiff = Math.max(mxDiff, mx - x);
            mx = Math.max(mx, x);
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long maximumTripletValue(vector<int>& nums) {
        long long ans = 0, mxDiff = 0;
        int mx = 0;
        for (int x : nums) {
            ans = max(ans, mxDiff * x);
            mxDiff = max(mxDiff, 1LL * mx - x);
            mx = max(mx, x);
        }
        return ans;
    }
};

Go

func maximumTripletValue(nums []int) int64 {
	ans, mx, mxDiff := 0, 0, 0
	for _, x := range nums {
		ans = max(ans, mxDiff*x)
		mxDiff = max(mxDiff, mx-x)
		mx = max(mx, x)
	}
	return int64(ans)
}

TypeScript

function maximumTripletValue(nums: number[]): number {
    let [ans, mx, mxDiff] = [0, 0, 0];
    for (const x of nums) {
        ans = Math.max(ans, mxDiff * x);
        mxDiff = Math.max(mxDiff, mx - x);
        mx = Math.max(mx, x);
    }
    return ans;
}

Rust

impl Solution {
    pub fn maximum_triplet_value(nums: Vec<i32>) -> i64 {
        let mut ans: i64 = 0;
        let mut mx: i32 = 0;
        let mut mx_diff: i32 = 0;

        for &x in &nums {
            ans = ans.max(mx_diff as i64 * x as i64);
            mx_diff = mx_diff.max(mx - x);
            mx = mx.max(x);
        }

        ans
    }
}