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

The optimal solution has a time complexity of O(n^3).

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

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

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

The brute-force approach of checking all possible triplets is straightforward but might be inefficient for larger arrays. No specific data structure is needed beyond iterating through the array. The problem can be solved using nested loops.

Which companies ask LeetCode 2873. Maximum Value of an Ordered Triplet I in interviews?

This problem is frequently asked at Media.net.

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Maximum Value of an Ordered Triplet I - LeetCode 2873 Solution

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

The provided code uses a brute-force approach. It iterates through all possible combinations of indices (i, j, k) such that i < j < k. For each triplet, it calculates the triplet value and updates the `max_value` if a larger value is found. Finally, it returns the `max_value`.

Maximum Value of an Ordered Triplet I - Complete Solution Guide

Maximum Value of an Ordered Triplet I is LeetCode problem 2873, a Easy 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 you to find the maximum value among all possible triplets (i, j, k) from a given array `nums`, where i < j < k. The value of a triplet is calculated as (nums[i] - nums[j]) * nums[k]. If all triplet values are negative, the function should return 0. The input is a 0-indexed integer array, and the output is a single integer representing the maximum triplet value.

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible combinations of indices (i, j, k) such that i < j < k. For each triplet, it calculates the triplet value and updates the `max_value` if a larger value is found. Finally, it returns the `max_value`.

Step-by-Step Algorithm

Initialize `max_value` to 0.
Iterate through the array using three nested loops to generate all possible triplets (i, j, k) where i < j < k.
For each triplet, calculate the value: `(nums[i] - nums[j]) * nums[k]`. Ensure that the calculation is performed with a data type that can handle potential overflows (e.g., `long long` in C++).
Update `max_value` if the calculated value is greater than the current `max_value`.
After checking all triplets, return `max_value`.

Key Insights

The brute-force approach of checking all possible triplets is straightforward but might be inefficient for larger arrays.
No specific data structure is needed beyond iterating through the array. The problem can be solved using nested loops.
The problem requires careful handling of potential integer overflow when multiplying numbers. Using a `long long` (or equivalent) data type helps prevent this.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Media.net.

Maximum Value of an Ordered Triplet I | LeetCode 2873

Easy

Array

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

Solution Code

Available in:

class Solution {
    public long maximumTripletValue(int[] nums) {
        long maxVal = 0;
        int n = nums.length;
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
                    long val = (long) (nums[i] - nums[j]) * nums[k];
                    maxVal = Math.max(maxVal, val);
                }
            }
        }
        return maxVal;
    }
}

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Problem Info

Description

Problem 2873: Maximum Value of an Ordered Triplet I - Detailed explanation and solution approaches.

Topics

Array

Companies

Media.net

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2873

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

5 steps

1
Initialize `max_value` to 0.
2
Iterate through the array using three nested loops to generate all possible triplets (i, j, k) where i < j < k.
3
For each triplet, calculate the value: `(nums[i] - nums[j]) * nums[k]`. Ensure that the calculation is performed with a data type that can handle potential overflows (e.g., `long long` in C++).
4
Update `max_value` if the calculated value is greater than the current `max_value`.
5
After checking all triplets, return `max_value`.

💡

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^3)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The three nested loops iterate through all possible triplets of indices. The number of such triplets is proportional to n^3, where n is the length of the array. The algorithm uses only a constant amount of extra space to store variables like `max_value` and loop counters.

📖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

✨

Key Insights

1The brute-force approach of checking all possible triplets is straightforward but might be inefficient for larger arrays.
2No specific data structure is needed beyond iterating through the array. The problem can be solved using nested loops.
3The problem requires careful handling of potential integer overflow when multiplying numbers. Using a `long long` (or equivalent) data type helps prevent this.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty array or array with fewer than 3 elements: The code handles this correctly because the outer loop condition `i < n - 2` prevents execution if the array is too small. The function returns 0 in this case, as there are no triplets.
All triplets have negative values: The code correctly returns 0 in this situation because `max_value` is initialized to 0, and if all calculated values are negative, `max_value` will remain 0.

⚠️

Common Mistakes to Avoid

✗Integer overflow: Failing to use a larger data type (like `long long` in C++ or `long` in Java) to store the intermediate calculation result can lead to incorrect answers due to integer overflow, especially when dealing with larger input values.

✗Incorrect loop bounds: Errors in the loop bounds (i, j, k) can result in missing potential triplets or accessing elements outside the array's boundaries, causing incorrect results or runtime errors.

✅

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?

💼Financial analysis: Finding maximum profit from a sequence of stock prices (where nums[i] represents the price on day i).

🏦Signal processing: Identifying maximum correlation or difference products between signal samples within a time window.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 21, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this easy-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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