What is the time complexity of Alternating Groups II?

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

What data structures are used to solve Alternating Groups II?

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

What are the key insights for solving Alternating Groups II?

Insight 1: To handle the circular nature of the problem efficiently, the `colors` array is extended by appending the first `k-1` elements to the end. This allows us to treat the circle as a linear array for easier sliding window traversal. Insight 2: Instead of re-checking the alternating condition for each window of size `k`, we maintain a count of 'alternating pairs' within the current window. By updating this count as the window slides, we can efficiently determine if the entire group is alternating.

Which companies ask LeetCode 3208. Alternating Groups II in interviews?

This problem is frequently asked at Samsara.

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Alternating Groups II - LeetCode 3208 Solution

Alternating Groups II - Complete Solution Guide

Alternating Groups II is LeetCode problem 3208, 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

There is a circle of red and blue tiles. You are given an array of integers colors and an integer k . 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 . An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles). Return the number of alternating groups. Note that since colors repr

Detailed Explanation

The problem asks us to find the number of alternating groups of size `k` in a circular arrangement of red (0) and blue (1) tiles. An alternating group is defined as a contiguous sequence of `k` tiles where each tile (except the first and last) has a different color from its immediate neighbors. Since it's a circle, the first and last tiles are also neighbors. The input is an array `colors` representing the tile colors and an integer `k` representing the group size. We need to return the total count of such alternating groups within the circle.

Solution Approach

The solution uses a sliding window approach. First, the input array `colors` is extended to simulate the circular nature by appending the first `k-1` elements to the end. Then, a sliding window of size `k` is moved across the extended array. For each window position, the solution checks if the tiles form an alternating group. To avoid repeatedly checking the alternating condition for each window, the solution maintains a counter, `alternating_pairs`, which stores the number of adjacent tiles in the window with different colors. As the window slides, `alternating_pairs` is updated by subtracting the alternating status of the exiting pair and adding the alternating status of the entering pair. If `alternating_pairs` equals `k-1`, then we have an alternating group.

Step-by-Step Algorithm

Step 1: Extend the `colors` array by appending the first `k-1` elements to simulate the circular arrangement.
Step 2: Initialize `alternating_pairs` to 0 and calculate the initial number of alternating pairs within the first window of size `k`.
Step 3: Initialize a `count` variable to 0 to keep track of the number of alternating groups.
Step 4: If the initial `alternating_pairs` is equal to `k-1`, increment the `count`.
Step 5: Iterate from the second window (index 1) to the end of the original `colors` array. In each iteration:
Step 6: Update `alternating_pairs` by subtracting 1 if the exiting pair (colors[i-1] and colors[i]) are different, and adding 1 if the entering pair (colors[i+k-2] and colors[i+k-1]) are different.
Step 7: If the updated `alternating_pairs` is equal to `k-1`, increment the `count`.
Step 8: Return the final `count`.

Key Insights

Insight 1: To handle the circular nature of the problem efficiently, the `colors` array is extended by appending the first `k-1` elements to the end. This allows us to treat the circle as a linear array for easier sliding window traversal.
Insight 2: Instead of re-checking the alternating condition for each window of size `k`, we maintain a count of 'alternating pairs' within the current window. By updating this count as the window slides, we can efficiently determine if the entire group is alternating.
Insight 3: The condition for an alternating group is that the number of alternating pairs within the k-sized group equals k-1. This is because in an alternating sequence of length k, there are k-1 pairs of adjacent elements, and each of these must be alternating.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Sliding Window.

Companies

Asked at: Samsara.

Alternating Groups II | LeetCode 3208

Medium

Array Sliding Window

Time: O(n)Space: O(k)

Solution Code

Available in:

class Solution {
    public int numberOfAlternatingGroups(int[] colors, int k) {
        int n = colors.length;
        int[] extended_colors = new int[n + k - 1];
        for (int i = 0; i < n; i++) {
            extended_colors[i] = colors[i];
        }
        for (int i = 0; i < k - 1; i++) {
            extended_colors[n + i] = colors[i];
        }
        
        int alternating_pairs = 0;
        for (int i = 0; i < k - 1; i++) {
            if (extended_colors[i] != extended_colors[i + 1]) {
                alternating_pairs += 1;
            }
        }
        
        int count = 0;
        if (alternating_pairs == k - 1) {
            count += 1;
        }
        
        for (int i = 1; i < n; i++) {
            if (extended_colors[i - 1] != extended_colors[i]) {
                alternating_pairs -= 1;
            }
            
            if (extended_colors[i + k - 2] != extended_colors[i + k - 1]) {
                alternating_pairs += 1;
            }
            
            if (alternating_pairs == k - 1) {
                count += 1;
            }
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 3208: Alternating Groups II - Detailed explanation and solution approaches.

Topics

Array Sliding Window

Companies

Samsara

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3208

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: Extend the `colors` array by appending the first `k-1` elements to simulate the circular arrangement.
2
Step 2: Initialize `alternating_pairs` to 0 and calculate the initial number of alternating pairs within the first window of size `k`.
3
Step 3: Initialize a `count` variable to 0 to keep track of the number of alternating groups.
4
Step 4: If the initial `alternating_pairs` is equal to `k-1`, increment the `count`.
5
Step 5: Iterate from the second window (index 1) to the end of the original `colors` array. In each iteration:
6
Step 6: Update `alternating_pairs` by subtracting 1 if the exiting pair (colors[i-1] and colors[i]) are different, and adding 1 if the entering pair (colors[i+k-2] and colors[i+k-1]) are different.
7
Step 7: If the updated `alternating_pairs` is equal to `k-1`, increment the `count`.
8
Step 8: Return the final `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(k)

📝

Why?

The time complexity is O(n) because the dominant operation is the sliding window traversal, which processes each element of the array once. The space complexity is O(k) due to the extended colors array using k-1 extra elements.

📖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

Iterate through all possible starting positions for a group of size `k`, and for each group, check if it's alternating by comparing adjacent tiles. The circular nature needs to be handled with modulo arithmetic.

Time: O(n*k), where n is the length of `colors` and k is the group size.Space: O(1). No extra space required.

Trade-offs: This approach is simpler to understand, but less efficient due to the nested loops involved in checking the alternating condition for each group. Use this approach only if performance is not a concern.

✨

Key Insights

1Insight 1: To handle the circular nature of the problem efficiently, the `colors` array is extended by appending the first `k-1` elements to the end. This allows us to treat the circle as a linear array for easier sliding window traversal.
2Insight 2: Instead of re-checking the alternating condition for each window of size `k`, we maintain a count of 'alternating pairs' within the current window. By updating this count as the window slides, we can efficiently determine if the entire group is alternating.
3Insight 3: The condition for an alternating group is that the number of alternating pairs within the k-sized group equals k-1. This is because in an alternating sequence of length k, there are k-1 pairs of adjacent elements, and each of these must be alternating.

📚

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: Input array with all same color tiles. In this case, `alternating_pairs` will always be 0, so the `count` will remain 0, which is the correct answer.
Edge case 2: Input array where colors alternate from beginning to end, like [0, 1, 0, 1, ...]. The sliding window logic correctly identifies all valid alternating groups.
Edge case 3: k = 3, colors = [0,1,0,1,0,1,0], Here we expect an alternating group at every index because any sequence of three tiles in this circular configuration is alternating. Solution handles this correctly

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the circular nature of the array correctly, which leads to incorrect results at the boundaries of the array. The extension strategy is crucial to solve this.

✗Common mistake 2: Repeatedly checking the alternating condition for each window, which results in O(n*k) time complexity. Optimizing the solution by keeping track of the number of alternating pairs provides better efficiency.

✗Common mistake 3: Off-by-one errors when indexing the `extended_colors` array, especially when updating the `alternating_pairs` count.

✅

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 network traffic patterns for alternating sequences of requests and responses to detect anomalies or attacks.

🏦Real-world scenario 2: Identifying alternating patterns in financial data, such as alternating periods of high and low trading volume, to predict market trends.

🎮Real-world scenario 3: Quality control in manufacturing, where alternating defects might indicate a systematic issue with the production process.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 27, 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 14

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

@TechLeadJul 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.

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

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