What is the time complexity of Alternating Groups I?

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

What data structures are used to solve Alternating Groups I?

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

What are the key insights for solving Alternating Groups I?

Insight 1: Circularity of the array needs to be handled correctly. We must consider the wrap-around effect when accessing the neighbors of the first and last elements. Insight 2: A simple iterative approach suffices. We can traverse the array and check the color of each tile and its neighbors to identify alternating groups.

Which companies ask LeetCode 3206. Alternating Groups I in interviews?

This problem is frequently asked at Samsara.

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Alternating Groups I - LeetCode 3206 Solution

Alternating Groups I - Complete Solution Guide

Alternating Groups I is LeetCode problem 3206, 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

There is a circle of red and blue tiles. You are given an array of integers colors . The color of tile i is represented by colors[i] : colors[i] == 0 means that tile i is red . colors[i] == 1 means that tile i is blue . Every 3 contiguous tiles in the circle with alternating colors (the middle tile has a different color from its left and right tiles) is called an alternating group. Return the number of alternating groups. Note that since colors represents a circle , the first and the last tiles

Detailed Explanation

The problem asks to count the number of 'alternating groups' in a circular array of colors. An alternating group is a sequence of three consecutive tiles where the middle tile's color is different from both its left and right neighbors. The array represents a circle, meaning the last element is considered adjacent to the first element. The input is an array of integers, where 0 represents red and 1 represents blue. The output is the number of alternating groups found.

Solution Approach

The solution uses a single loop to iterate through the `colors` array. For each element, it checks its left and right neighbors using the modulo operator to handle the circularity. If the current element's color is different from both its left and right neighbors, it's considered part of an alternating group, and the `count` is incremented.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 0.
Step 2: Iterate through the `colors` array using a `for` loop.
Step 3: For each element `colors[i]`, calculate the indices of its left and right neighbors using the modulo operator: `prev = (i - 1 + n) % n` and `next = (i + 1) % n`, where `n` is the length of the array.
Step 4: Check if `colors[prev] != colors[i]` and `colors[i] != colors[next]`. If this condition is true, increment `count`.
Step 5: After iterating through all elements, return `count`.

Key Insights

Insight 1: Circularity of the array needs to be handled correctly. We must consider the wrap-around effect when accessing the neighbors of the first and last elements.
Insight 2: A simple iterative approach suffices. We can traverse the array and check the color of each tile and its neighbors to identify alternating groups.
Insight 3: The modulo operator (%) is crucial for efficiently handling the circular nature of the array. It ensures that when we try to access elements before the beginning or after the end of the array, we get a valid index.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Sliding Window.

Companies

Asked at: Samsara.

Alternating Groups I | LeetCode 3206

Easy

Array Sliding Window

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

Solution Code

Available in:

class Solution {
    public int numberOfAlternatingGroups(int[] colors) {
        int n = colors.length;
        int count = 0;
        for (int i = 0; i < n; i++) {
            int prev = (i - 1 + n) % n;
            int next = (i + 1) % n;
            if (colors[prev] != colors[i] && colors[i] != colors[next]) {
                count++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 3206: Alternating Groups I - Detailed explanation and solution approaches.

Topics

Array Sliding Window

Companies

Samsara

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3206

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
Step 1: Initialize a `count` variable to 0.
2
Step 2: Iterate through the `colors` array using a `for` loop.
3
Step 3: For each element `colors[i]`, calculate the indices of its left and right neighbors using the modulo operator: `prev = (i - 1 + n) % n` and `next = (i + 1) % n`, where `n` is the length of the array.
4
Step 4: Check if `colors[prev] != colors[i]` and `colors[i] != colors[next]`. If this condition is true, increment `count`.
5
Step 5: After iterating through all elements, 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution iterates through the array once, performing a constant number of operations for each element. Therefore, the time complexity is linear, O(n), where n is the length of the `colors` array. The space complexity is constant because it uses only a few integer variables regardless of the array 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

✨

Key Insights

1Insight 1: Circularity of the array needs to be handled correctly. We must consider the wrap-around effect when accessing the neighbors of the first and last elements.
2Insight 2: A simple iterative approach suffices. We can traverse the array and check the color of each tile and its neighbors to identify alternating groups.
3Insight 3: The modulo operator (%) is crucial for efficiently handling the circular nature of the array. It ensures that when we try to access elements before the beginning or after the end of the array, we get a valid index.

📚

Key Terms Explained

For Students

Array

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

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: An array with all the same colors. The solution correctly handles this by returning 0 because no alternating groups exist.
Edge case 2: An array with only three elements. The solution accurately checks if these three elements form an alternating group.
Edge case 3: The circular nature of the array. The solution correctly handles the wrapping around from the last element to the first and vice versa.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the circularity of the array. Beginners might forget to use the modulo operator when accessing neighbors, leading to incorrect results for the first and last elements.

✗Common mistake 2: Off-by-one errors in index calculations. Care must be taken to correctly calculate the indices of the neighbors to avoid accessing elements out of bounds or missing alternating groups.

✅

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 patterns in sequences of events, where the pattern is defined by alternating states. For example, analyzing traffic patterns (stop-go), or machine cycles.

🏦Real-world scenario 2: Image processing, where the algorithm could identify patterns of alternating pixels. For example, detecting edges in a binary image where black and white pixels alternate.

💡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: October 25, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

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 "Alternating Groups 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 sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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