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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Value of an Ordered Triplet II?

This problem uses Array. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Value of an Ordered Triplet II?

Insight 1: The key to solving this problem efficiently is to realize that we don't need to generate all possible triplets explicitly. Instead, we can iterate through the array once, keeping track of the maximum value of `nums[i]` seen so far and the maximum value of `nums[i] - nums[j]` seen so far, where i < j. Insight 2: The problem can be solved with a single pass through the array, updating `max_i` and `max_ij_diff` dynamically. This avoids the O(n^3) complexity of brute-force triplet generation.

Which companies ask LeetCode 2874. Maximum Value of an Ordered Triplet II in interviews?

This problem is frequently asked at Media.net.

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Maximum Value of an Ordered Triplet II - LeetCode 2874 Solution

Q: How to solve Maximum Value of an Ordered Triplet II?

The solution iterates through the `nums` array once. At each element `nums[k]`, it calculates the value of the triplet ending at `k` using previously computed `max_ij_diff`. The maximum result is updated accordingly. `max_i` keeps track of the maximum number encountered so far, and `max_ij_diff` maintains the maximum difference `nums[i] - nums[j]` encountered so far. This way, we only iterate once and only keep a few variables, resulting in a linear time solution.

Q: Which companies ask LeetCode 2874. Maximum Value of an Ordered Triplet II in interviews?

This problem is frequently asked at Media.net.

Maximum Value of an Ordered Triplet II - Complete Solution Guide

Maximum Value of an Ordered Triplet II is LeetCode problem 2874, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to find the maximum value among all possible ordered triplets (i, j, k) in a given array `nums`, where i < j < k. The value of a triplet is calculated as (nums[i] - nums[j]) * nums[k]. If all possible triplets have a negative value, the function should return 0. The input is an array of integers, and the output is the maximum triplet value (or 0 if all are negative). The constraints specify that the array length is between 3 and 10^5, and the numbers in the array are between 1 and 10^6.

Solution Approach

The solution iterates through the `nums` array once. At each element `nums[k]`, it calculates the value of the triplet ending at `k` using previously computed `max_ij_diff`. The maximum result is updated accordingly. `max_i` keeps track of the maximum number encountered so far, and `max_ij_diff` maintains the maximum difference `nums[i] - nums[j]` encountered so far. This way, we only iterate once and only keep a few variables, resulting in a linear time solution.

Step-by-Step Algorithm

Step 1: Initialize `res` to 0. This variable will store the maximum triplet value found so far.
Step 2: Initialize `max_i` to 0. This variable will store the maximum value of `nums[i]` encountered so far.
Step 3: Initialize `max_ij_diff` to 0. This variable will store the maximum value of `nums[i] - nums[j]` encountered so far.
Step 4: Iterate through the `nums` array. In each iteration (let `num` be the current element):
Step 5: Update `res` with the maximum of its current value and `max_ij_diff * num`. This represents the maximum triplet value seen so far with `num` being the last element of the triplet.
Step 6: Update `max_ij_diff` with the maximum of its current value and `max_i - num`. This ensures we maintain the maximum possible difference between a previous element and the current element.
Step 7: Update `max_i` with the maximum of its current value and `num`. This maintains the maximum element seen so far.
Step 8: After iterating through the entire array, return `res`.

Key Insights

Insight 1: The key to solving this problem efficiently is to realize that we don't need to generate all possible triplets explicitly. Instead, we can iterate through the array once, keeping track of the maximum value of `nums[i]` seen so far and the maximum value of `nums[i] - nums[j]` seen so far, where i < j.
Insight 2: The problem can be solved with a single pass through the array, updating `max_i` and `max_ij_diff` dynamically. This avoids the O(n^3) complexity of brute-force triplet generation.
Insight 3: Since the values in the array and intermediate results can be large, using `long long` (or `long` in Java) for calculations is crucial to prevent integer overflow. The result may be zero, which is handled appropriately.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Media.net.

Maximum Value of an Ordered Triplet II | LeetCode 2874

Medium

Array

Time: O(n)Space: O(1)

Solution Code

Available in:

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

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Problem Info

Description

Problem 2874: Maximum Value of an Ordered Triplet II - Detailed explanation and solution approaches.

Topics

Array

Companies

Media.net

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2874

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `res` to 0. This variable will store the maximum triplet value found so far.
2
Step 2: Initialize `max_i` to 0. This variable will store the maximum value of `nums[i]` encountered so far.
3
Step 3: Initialize `max_ij_diff` to 0. This variable will store the maximum value of `nums[i] - nums[j]` encountered so far.
4
Step 4: Iterate through the `nums` array. In each iteration (let `num` be the current element):
5
Step 5: Update `res` with the maximum of its current value and `max_ij_diff * num`. This represents the maximum triplet value seen so far with `num` being the last element of the triplet.
6
Step 6: Update `max_ij_diff` with the maximum of its current value and `max_i - num`. This ensures we maintain the maximum possible difference between a previous element and the current element.
7
Step 7: Update `max_i` with the maximum of its current value and `num`. This maintains the maximum element seen so far.
8
Step 8: After iterating through the entire array, return `res`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we loop through the array only once. The space complexity is O(1) as we are only using a few variables to store temporary results and not using any extra data structures dependent on the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Approach

Iterate through all possible triplets (i, j, k) with i < j < k, calculate the value of each triplet, and keep track of the maximum value seen so far.

Time: O(n^3)Space: O(1)

Trade-offs: The brute-force approach is very simple to understand but is inefficient for large input arrays because of its cubic time complexity. It is only suitable for very small input sizes. The optimized approach is much faster for any reasonable input size.

✨

Key Insights

1Insight 1: The key to solving this problem efficiently is to realize that we don't need to generate all possible triplets explicitly. Instead, we can iterate through the array once, keeping track of the maximum value of `nums[i]` seen so far and the maximum value of `nums[i] - nums[j]` seen so far, where i < j.
2Insight 2: The problem can be solved with a single pass through the array, updating `max_i` and `max_ij_diff` dynamically. This avoids the O(n^3) complexity of brute-force triplet generation.
3Insight 3: Since the values in the array and intermediate results can be large, using `long long` (or `long` in Java) for calculations is crucial to prevent integer overflow. The result may be zero, which is handled appropriately.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: All elements are the same. The algorithm handles this correctly because if all elements are equal, `max_ij_diff` will eventually become 0, and the result will be 0.
Edge case 2: The array is in descending order. Even if the difference `nums[i] - nums[j]` is negative, the algorithm correctly tracks `max_ij_diff` and will find the maximum (which might be 0 if all triplet values are negative).
Edge case 3: Integer overflow. Using `long long` in C++ and C, and `long` in Java is important to prevent integer overflow since the product of (nums[i] - nums[j]) * nums[k] can exceed the range of a regular integer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using integers instead of long longs/longs, leading to potential integer overflow when calculating the triplet value.

✗Common mistake 2: Trying to pre-calculate all possible differences (nums[i] - nums[j]) which results in higher time complexity and often unnecessary storage.

✗Common mistake 3: Incorrectly implementing the loop conditions or comparisons, leading to incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing financial data to identify maximum profit potential from buying and selling stocks over a period of time. Here, the array can represent stock prices, and finding the maximum triplet value might represent identifying the best buy-sell-buy strategy.

🏦Real-world scenario 2: Optimizing resource allocation in a supply chain. The array can represent costs or profits at different stages, and finding the maximum triplet value might help in identifying the most profitable combination of resource allocation strategies.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Value of an Ordered Triplet II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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