LeetCode 2873: Maximum Value of an Ordered Triplet I Solution

Master LeetCode problem 2873 (Maximum Value of an Ordered Triplet I), a easy 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.

2873. Maximum Value of an Ordered Triplet I

Easy

Array

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible triplets of indices (i, j, k) where i < j < k. For each triplet, we calculate the value as (nums[i] - nums[j]) * nums[k] and keep track of the maximum value found. If all such triplets have a negative value, we return 0 as the result.

Algorithm:

Initialize a variable maxValue to keep track of the maximum value found.
Iterate through all possible triplets (i, j, k) where i < j < k.
Calculate the value as (nums[i] - nums[j]) * nums[k].
Update maxValue if the calculated value is greater than the current maxValue.
If all triplets have negative values, return 0 as the result.

Time Complexity:

The time complexity of this algorithm is O(n^3), where n is the length of the input array nums.

Space Complexity:

The space complexity of this algorithm is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int maxTripletValue(int[] nums) {
        int maxValue = 0;
        for (int i = 0; i < nums.length - 2; i++) {
            for (int j = i + 1; j < nums.length - 1; j++) {
                for (int k = j + 1; k < nums.length; k++) {
                    int value = (nums[i] - nums[j]) * nums[k];
                    maxValue = Math.max(maxValue, value);
                }
            }
        }
        return maxValue;
    }
}

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