What is the time complexity of Count Alternating Subarrays?

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

What data structures are used to solve Count Alternating Subarrays?

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

What are the key insights for solving Count Alternating Subarrays?

Insight 1: Each single element is considered an alternating subarray. Insight 2: We can efficiently count alternating subarrays by iterating through the array and maintaining a 'streak' of alternating elements.

Which companies ask LeetCode 3101. Count Alternating Subarrays in interviews?

This problem is frequently asked at Capital One.

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Count Alternating Subarrays - LeetCode 3101 Solution

Count Alternating Subarrays - Complete Solution Guide

Count Alternating Subarrays is LeetCode problem 3101, 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 binary array nums . We call a subarray alternating if no two adjacent elements in the subarray have the same value. Return the number of alternating subarrays in nums . Example 1: Input: nums = [0,1,1,1] Output: 5 Explanation: The following subarrays are alternating: [0] , [1] , [1] , [1] , and [0,1] . Example 2: Input: nums = [1,0,1,0] Output: 10 Explanation: Every subarray of the array is alternating. There are 10 possible subarrays that we can choose. Constraints: 1 <= nums.le

Detailed Explanation

The problem asks us to count the number of alternating subarrays within a given binary array. An alternating subarray is defined as a subarray where no two adjacent elements have the same value (e.g., [0, 1, 0, 1] is alternating, but [0, 1, 1, 0] is not). The input is a binary array `nums`, and the output is the total number of alternating subarrays present within it. The constraints specify that the array length is between 1 and 10<sup>5</sup>, and each element is either 0 or 1.

Solution Approach

The provided solution uses a single pass through the input array `nums` to count the alternating subarrays. It maintains a `current_streak` variable to track the length of the current alternating sequence. If the current element is different from the previous one, the streak is incremented. Otherwise, the streak is reset to 1. The `total_count` is updated by adding the `current_streak` at each iteration. This works because at each index `i`, the `current_streak` represents the number of alternating subarrays that *end* at index `i`. Summing this `current_streak` over all indices effectively counts all alternating subarrays within the input array.

Step-by-Step Algorithm

Step 1: Initialize `total_count` and `current_streak` to 0.
Step 2: Iterate through the `nums` array from left to right (index `i` from 0 to `len(nums) - 1`).
Step 3: Inside the loop, check if `i > 0` (to avoid out-of-bounds access) and if `nums[i]` is different from `nums[i-1]`.
Step 4: If `nums[i] != nums[i-1]`, increment `current_streak` by 1 because we're extending the alternating sequence.
Step 5: Otherwise (if `nums[i] == nums[i-1]` or `i == 0`), reset `current_streak` to 1 because a new alternating sequence starts at index `i`.
Step 6: Add the `current_streak` to `total_count`. This counts all alternating subarrays ending at index `i`.
Step 7: After the loop completes, return `total_count`.

Key Insights

Insight 1: Each single element is considered an alternating subarray.
Insight 2: We can efficiently count alternating subarrays by iterating through the array and maintaining a 'streak' of alternating elements.
Insight 3: The number of alternating subarrays ending at a given index 'i' is equal to the length of the current alternating streak ending at that index.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: Capital One.

Count Alternating Subarrays | LeetCode 3101

Medium

Array Math

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

Solution Code

Available in:

class Solution {
    public long countAlternatingSubarrays(int[] nums) {
        long total_count = 0;
        int current_streak = 0;
        
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] != nums[i-1]) {
                current_streak += 1;
            } else {
                current_streak = 1;
            }
            
            total_count += current_streak;
        }
        
        return total_count;
    }
}

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Problem Info

Description

Problem 3101: Count Alternating Subarrays - Detailed explanation and solution approaches.

Topics

Array Math

Companies

Capital One

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3101

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
Step 1: Initialize `total_count` and `current_streak` to 0.
2
Step 2: Iterate through the `nums` array from left to right (index `i` from 0 to `len(nums) - 1`).
3
Step 3: Inside the loop, check if `i > 0` (to avoid out-of-bounds access) and if `nums[i]` is different from `nums[i-1]`.
4
Step 4: If `nums[i] != nums[i-1]`, increment `current_streak` by 1 because we're extending the alternating sequence.
5
Step 5: Otherwise (if `nums[i] == nums[i-1]` or `i == 0`), reset `current_streak` to 1 because a new alternating sequence starts at index `i`.
6
Step 6: Add the `current_streak` to `total_count`. This counts all alternating subarrays ending at index `i`.
7
Step 7: After the loop completes, return `total_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 time complexity is linear because each element in the array is visited and processed exactly once. The space complexity is constant because the amount of extra memory used doesn't depend on the input size; it's always the same few 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

Brute-force approach

Generate all possible subarrays and check if each one is alternating. If it is, increment a counter.

Time: O(n^3) or O(n^2) depending on the alternating sequence check implementation (O(n) or O(1) per subarray, respectively, and O(n^2) subarrays).Space: O(1)

Trade-offs: This approach is simple to understand but highly inefficient. It should only be used for very small input sizes, as the O(n^2) or O(n^3) time complexity makes it impractical for larger arrays. The main solution is much faster with O(n) time complexity. The brute-force approach checks every subarray, while the optimized one calculates the number of alternating arrays dynamically.

✨

Key Insights

1Insight 1: Each single element is considered an alternating subarray.
2Insight 2: We can efficiently count alternating subarrays by iterating through the array and maintaining a 'streak' of alternating elements.
3Insight 3: The number of alternating subarrays ending at a given index 'i' is equal to the length of the current alternating streak ending at that index.

📚

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: `nums` is empty. The problem states 1 <= nums.length, so this case is not possible. However, if it were, the code would correctly return 0 due to the initial value of `total_count`.
Edge case 2: `nums` contains only one element (e.g., `[0]`). The code correctly handles this case because `current_streak` is initialized to 0, set to 1 in the first iteration, and added to `total_count`, resulting in a final count of 1.
Edge case 3: `nums` contains all the same elements (e.g., `[0, 0, 0]`). The `current_streak` will always be 1, so the `total_count` will be the length of `nums` (which is correct).
Edge case 4: `nums` contains a perfect alternating sequence (e.g., `[0, 1, 0, 1]`). The `current_streak` will increment by 1 in each iteration after the first, leading to a `total_count` that reflects the sum of the sequence 1 + 2 + 3 + ... + n (which is also correct).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering single-element subarrays as alternating. Remember that each element itself forms a valid alternating subarray.

✗Common mistake 2: Incorrectly resetting the `current_streak`. The streak should only be reset when two adjacent elements are the same.

✗Common mistake 3: Using nested loops to generate and check all subarrays, leading to a higher time complexity. The problem can be solved with a single pass.

✗Common mistake 4: Integer overflow. While the problem does not specifically state the necessity of using a long type, the count of subarrays can become very large for large arrays, and this could lead to integer overflow. The Java and C++ solutions, along with C, are correctly using `long` or `long long` to prevent this issue. Although the Python solution does not have a maximum integer size, it is generally considered good practice to be mindful of possible overflow cases.

✅

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: Signal processing. Identifying alternating patterns in a signal stream can be used for noise reduction or feature extraction.

🏦Real-world scenario 2: Data compression. Alternating sequences can be efficiently compressed using run-length encoding or similar techniques.

🎮Real-world scenario 3: Pattern recognition. Identifying alternating patterns in financial data, such as stock prices or trading volumes, can be used for trend analysis or anomaly detection.

📱Real-world scenario 4: DNA Sequencing. Analyzing DNA sequences to identify alternating patterns of base pairs, which can be indicative of certain genetic markers or mutations.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 12, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Count Alternating Subarrays"?

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

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