How do you solve Maximum Value of an Ordered Triplet I on LeetCode?

Maximum Value of an Ordered Triplet I (LeetCode #2873) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The value of the triplet is influenced heavily by the choice of k, as it multiplies the difference between nums[i] and nums[j].

What is the time complexity of Maximum Value of an Ordered Triplet I?

The optimal solution for Maximum Value of an Ordered Triplet I runs in O(n²) time and uses O(n) space. The time complexity is O(n²) due to the two nested loops, while the space complexity is O(n) for the max_k array.

What is the brute force solution for Maximum Value of an Ordered Triplet I?

The brute force approach for Maximum Value of an Ordered Triplet I is Brute Force, with O(n³) time complexity and O(1) space complexity. The brute force approach checks all possible triplets of indices (i, j, k) to calculate their values. This method is straightforward but inefficient, as it examines every combination. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Value of an Ordered Triplet I?

Maximum Value of an Ordered Triplet I uses the following concepts: Array. The recommended patterns to study are: Array, Dynamic Programming.

What are the interview tips for Maximum Value of an Ordered Triplet I?

Always clarify the constraints and edge cases before jumping into coding. Consider both brute force and optimized solutions; explain your thought process clearly. Dry run your algorithm with sample inputs to ensure correctness before finalizing your code.

What are common mistakes when solving Maximum Value of an Ordered Triplet I?

Not considering the case where all triplet values are negative and returning 0. Failing to maintain the order of indices (i < j < k) when iterating.

#2873

Maximum Value of an Ordered Triplet I

Easy

ArrayArrayDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

Instead of checking all triplets, we can optimize our approach by precomputing the maximum values for nums[k] and using them to calculate the maximum triplet value efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize max_value to 0 and create an array max_k to store maximum values of nums[k] for each index.
2Step 2: Fill max_k such that max_k[k] contains the maximum value of nums from index k to the end.
3Step 3: Iterate through the array with two loops to find i and j, and use max_k to compute the triplet value efficiently.

solution.py12 lines

1# Full working Python code
2nums = [12, 6, 1, 2, 7]
3max_k = [0] * len(nums)
4max_k[-1] = nums[-1]
5for i in range(len(nums) - 2, -1, -1):
6    max_k[i] = max(max_k[i + 1], nums[i])
7max_value = 0
8for i in range(len(nums) - 2):
9    for j in range(i + 1, len(nums) - 1):
10        value = (nums[i] - nums[j]) * max_k[j + 1]
11        max_value = max(max_value, value)
12print(max_value)

ℹ

Complexity note: The time complexity is O(n²) due to the two nested loops, while the space complexity is O(n) for the max_k array.

1The value of the triplet is influenced heavily by the choice of k, as it multiplies the difference between nums[i] and nums[j].
2Maximizing nums[k] while minimizing nums[j] relative to nums[i] is key to maximizing the triplet value.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.