What is the time complexity of Count Good Triplets?

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

What data structures are used to solve Count Good Triplets?

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

What are the key insights for solving Count Good Triplets?

The problem requires checking all possible triplets, leading to a nested loop structure. Brute-force iteration is a straightforward approach, but can be computationally expensive for large arrays.

Which companies ask LeetCode 1534. Count Good Triplets in interviews?

This problem is frequently asked at Turvo.

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Count Good Triplets - LeetCode 1534 Solution

Q: How to solve Count Good Triplets?

The provided solutions use a brute-force approach. They iterate through all possible triplets (i, j, k) using three nested loops. For each triplet, they check if the three conditions for a 'good triplet' are met. If all conditions are true, a counter is incremented. Finally, the counter (representing the total number of good triplets) is returned.

Count Good Triplets - Complete Solution Guide

Count Good Triplets is LeetCode problem 1534, 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

Given an array of integers arr , and three integers a , b and c . You need to find the number of good triplets. A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true: 0 <= i < j < k < arr.length |arr[i] - arr[j]| <= a |arr[j] - arr[k]| <= b |arr[i] - arr[k]| <= c Where |x| denotes the absolute value of x . Return the number of good triplets . Example 1: Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3 Output: 4 Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (

Detailed Explanation

The problem asks you to count the number of "good triplets" in an array of integers. A triplet (arr[i], arr[j], arr[k]) is considered good if it satisfies three conditions: 1) The indices i, j, and k are in strictly increasing order (0 <= i < j < k < arr.length). 2) The absolute differences between consecutive elements in the triplet are less than or equal to specified thresholds (|arr[i] - arr[j]| <= a, |arr[j] - arr[k]| <= b). 3) The absolute difference between the first and last elements of the triplet is also less than or equal to a specified threshold (|arr[i] - arr[k]| <= c). The input includes the array `arr` and the thresholds `a`, `b`, and `c`. The output is the total count of good triplets.

Solution Approach

The provided solutions use a brute-force approach. They iterate through all possible triplets (i, j, k) using three nested loops. For each triplet, they check if the three conditions for a 'good triplet' are met. If all conditions are true, a counter is incremented. Finally, the counter (representing the total number of good triplets) is returned.

Step-by-Step Algorithm

Initialize a counter `count` to 0.
Iterate through all possible values of `i` from 0 to `arr.length - 3`.
For each `i`, iterate through all possible values of `j` from `i + 1` to `arr.length - 2`.
For each pair `(i, j)`, iterate through all possible values of `k` from `j + 1` to `arr.length - 1`.
For each triplet `(i, j, k)`, check if the three conditions (`|arr[i] - arr[j]| <= a`, `|arr[j] - arr[k]| <= b`, `|arr[i] - arr[k]| <= c`) are satisfied.
If all three conditions are true, increment `count`.
After iterating through all triplets, return `count`.

Key Insights

The problem requires checking all possible triplets, leading to a nested loop structure.
Brute-force iteration is a straightforward approach, but can be computationally expensive for large arrays.
No significant optimization is possible beyond careful coding and using efficient built-in functions for absolute value calculation.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(1)

Topics

This problem involves: Array, Enumeration.

Companies

Asked at: Turvo.

Count Good Triplets | LeetCode 1534

Easy

Array Enumeration

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

Solution Code

Available in:

class Solution {
    public int countGoodTriplets(int[] arr, int a, int b, int c) {
        int count = 0;
        for (int i = 0; i < arr.length - 2; i++) {
            for (int j = i + 1; j < arr.length - 1; j++) {
                for (int k = j + 1; k < arr.length; k++) {
                    if (Math.abs(arr[i] - arr[j]) <= a && Math.abs(arr[j] - arr[k]) <= b && Math.abs(arr[i] - arr[k]) <= c) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 1534: Count Good Triplets - Detailed explanation and solution approaches.

Topics

Array Enumeration

Companies

Turvo

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1534

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

7 steps

1
Initialize a counter `count` to 0.
2
Iterate through all possible values of `i` from 0 to `arr.length - 3`.
3
For each `i`, iterate through all possible values of `j` from `i + 1` to `arr.length - 2`.
4
For each pair `(i, j)`, iterate through all possible values of `k` from `j + 1` to `arr.length - 1`.
5
For each triplet `(i, j, k)`, check if the three conditions (`|arr[i] - arr[j]| <= a`, `|arr[j] - arr[k]| <= b`, `|arr[i] - arr[k]| <= c`) are satisfied.
6
If all three conditions are true, increment `count`.
7
After iterating through all triplets, return `count`.

💡

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 time complexity is O(n^3) because of the three nested loops, where n is the length of the input array. Each loop iterates up to n times in the worst case. The space complexity is O(1) because the algorithm uses only a constant amount of extra space to store the counter and intermediate variables.

📖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 problem requires checking all possible triplets, leading to a nested loop structure.
2Brute-force iteration is a straightforward approach, but can be computationally expensive for large arrays.
3No significant optimization is possible beyond careful coding and using efficient built-in functions for absolute value calculation.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty array: The code handles the empty array case correctly; the outer loop won't execute, resulting in a count of 0.
Array with less than 3 elements: The outer loop condition prevents execution, resulting in 0, which is correct as you can't form a triplet.

⚠️

Common Mistakes to Avoid

✗Incorrectly implementing the nested loop conditions: Off-by-one errors in loop bounds are common. Careful attention to the loop indices (i, j, k) is needed to ensure all triplets are considered and the conditions are correctly checked.

✗Inefficient absolute value calculation: While not a significant performance bottleneck in this case, using a less efficient method for calculating absolute values could slightly impact performance in extremely large datasets.

✅

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?

💼Data analysis: Identifying sets of data points that meet specific proximity criteria. For example, finding clusters of similar values in sensor readings.

🏦Pattern recognition: Detecting patterns in time-series data. For instance, finding sequences of events that occur within specific time intervals.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: September 3, 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 13

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

@DevExpert92Jul 6

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Count Good Triplets"?

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 enumeration techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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