Leetcode 2873: Maximum Value of an Ordered Triplet I

Input: The input consists of an integer array nums where 3 <= nums.length <= 100, and 1 <= nums[i] <= 10^6.

Example: nums = [10, 5, 3, 6, 9]

Constraints:

• 3 <= nums.length <= 100

• 1 <= nums[i] <= 10^6

Output: Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all values are negative, return 0.

Example: For input nums = [10, 5, 3, 6, 9], the output is 48.

Constraints:

Goal: The goal is to maximize the value of (nums[i] - nums[j]) * nums[k] for triplets (i, j, k) where i < j < k.

Steps:

• Iterate through each element in the array.

• For each element, calculate the value of all possible triplets using it as i, j, or k.

• Track the maximum value found during the iteration.

• Return the maximum value or 0 if no non-negative triplet value exists.

Goal: The solution must handle arrays with lengths up to 100 and values up to 1 million efficiently.

Steps:

• 3 <= nums.length <= 100

• 1 <= nums[i] <= 10^6

Assumptions:

• The array has at least 3 elements.

• Input: For input nums = [10, 5, 3, 6, 9], the output is 48.

• Explanation: The triplet (0, 2, 4) results in the maximum value (10 - 3) * 9 = 48.

Approach: To solve this problem, we need to iterate through all possible triplets and calculate their values. The maximum value can then be returned.

Observations:

• The problem requires examining every triplet (i, j, k).

• The values of the triplets depend on the subtraction of nums[i] and nums[j] multiplied by nums[k].

• Instead of checking all triplets brute force, we can iterate through the array while keeping track of the maximum possible values efficiently.

Steps:

• Initialize variables to keep track of the maximum values.

• For each element in the array, calculate the potential value for all possible triplets with that element.

• Update the maximum value whenever a higher value is found.

• Return the maximum value or 0 if no valid triplet exists.

Empty Inputs:

• The input will always have at least 3 elements, so no need to handle empty inputs.

Large Inputs:

• Ensure that the solution efficiently handles arrays with up to 100 elements.

Special Values:

• Consider cases where no valid triplet results in a non-negative value.

Constraints:

• The algorithm must run efficiently even for the largest inputs.

long long maximumTripletValue(vector<int>& nums) {
    long res = 0, mxa = 0, mxab = 0;
    for(long a: nums) {
        res = max(res, mxab * a);
        mxab = max(mxab, mxa - a);
        mxa = max(mxa, a);
    }
    return res;
}

1 : Function Declaration

long long maximumTripletValue(vector<int>& nums) {

Function declaration that accepts a vector of integers and returns a long integer, representing the maximum triplet value.

2 : Variable Initialization

    long res = 0, mxa = 0, mxab = 0;

Initialize three variables: 'res' to store the result, 'mxa' for the maximum value encountered so far, and 'mxab' for the maximum value of (mxa - a).

3 : Loop Setup

    for(long a: nums) {

Start a loop to iterate over each element 'a' in the input array 'nums'.

4 : Max Calculation

        res = max(res, mxab * a);

Update 'res' by comparing its current value with the product of 'mxab' and 'a', ensuring that the largest product is kept.

5 : Update mxab

        mxab = max(mxab, mxa - a);

Update 'mxab' to store the maximum value between its current value and the difference between 'mxa' and 'a'.

6 : Update mxa

        mxa = max(mxa, a);

Update 'mxa' to store the maximum value encountered so far during the loop.

7 : Return Result

    return res;

Return the final result, which is the maximum triplet value computed.

Best Case: O(n)

Average Case: O(n^2)

Worst Case: O(n^2)

Description: The time complexity is O(n^2) because we are considering all possible triplets.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is O(1) as we use a constant amount of extra space.

Leetcode 2873: Maximum Value of an Ordered Triplet I

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