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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: Maintain Maximum Prefix Value and Maximum Difference

We can use two variables $mx$ and $mx_diff$ to maintain the maximum prefix value and maximum difference, respectively. When traversing the array, we update these two variables, and the answer is the maximum value of all $mx_diff \times nums[i]$.

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

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

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

  • class Solution:
    def maximumTripletValue(self, nums: List[int]) -> int:
        ans = mx = mx_diff = 0
        for num in nums:
            ans = max(ans, mx_diff * num)
            mx = max(mx, num)
            mx_diff = max(mx_diff, mx - num)
        return ans


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

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

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