LeetCode 2874: Maximum Value of an Ordered Triplet II Solution

Master LeetCode problem 2874 (Maximum Value of an Ordered Triplet II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2874. Maximum Value of an Ordered Triplet II

Medium

Array

Problem Explanation

Explanation

To find the maximum value of an ordered triplet, we can iterate through all triplets of indices (i, j, k) where i < j < k, calculate the value of each triplet, and keep track of the maximum value found. We can do this by using three nested loops. If all triplets have negative values, we return 0.

Initialize a variable maxValue to store the maximum value found.
Iterate through all triplets (i, j, k) where i < j < k.
Calculate the value of the current triplet as (nums[i] - nums[j]) * nums[k].
Update maxValue if the calculated value is greater.
If after iterating through all triplets, maxValue is still negative, return 0. Otherwise, return maxValue.

Time complexity: O(n^3) where n is the length of the input array nums. Space complexity: O(1)

Solution Code

class Solution {
    public int maxTripletValue(int[] nums) {
        int n = nums.length;
        int maxValue = Integer.MIN_VALUE;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
                    int value = (nums[i] - nums[j]) * nums[k];
                    maxValue = Math.max(maxValue, value);
                }
            }
        }
        
        return maxValue > 0 ? maxValue : 0;
    }
}

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