What is the time complexity of Count Subarrays of Length Three With a Condition?

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

What data structures are used to solve Count Subarrays of Length Three With a Condition?

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

What are the key insights for solving Count Subarrays of Length Three With a Condition?

Insight 1: The problem can be solved efficiently using a single loop iterating through the array. We only need to check triplets of consecutive numbers. Insight 2: The core logic involves checking the condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` for each possible subarray of length 3. Careful consideration of integer division is needed to ensure correctness.

Which companies ask LeetCode 3392. Count Subarrays of Length Three With a Condition in interviews?

This problem is frequently asked at Cognizant.

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Count Subarrays of Length Three With a Condition - LeetCode 3392 Solution

Count Subarrays of Length Three With a Condition - Complete Solution Guide

Count Subarrays of Length Three With a Condition is LeetCode problem 3392, 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 integer array nums , return the number of subarrays of length 3 such that the sum of the first and third numbers equals exactly half of the second number. Example 1: Input: nums = [1,2,1,4,1] Output: 1 Explanation: Only the subarray [1,4,1] contains exactly 3 elements where the sum of the first and third numbers equals half the middle number. Example 2: Input: nums = [1,1,1] Output: 0 Explanation: [1,1,1] is the only subarray of length 3. However, its first and third numbers do not add

Detailed Explanation

The problem asks you to count the number of subarrays of length 3 within a given integer array `nums` that satisfy a specific condition. The condition is that the sum of the first and third elements of the subarray must be exactly half of the second element. For example, in the subarray [1, 4, 1], 1 + 1 = 2, and 2 is half of 4, so this subarray satisfies the condition. The input is an integer array, and the output is the count of such subarrays.

Solution Approach

The solution uses a single loop to iterate through the input array `nums`. For each element at index `i`, it checks if a subarray of length 3 starting at `i` satisfies the given condition. The condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` is evaluated. If true, the `count` is incremented. The loop continues until the end of the array is reached, ensuring all possible subarrays of length 3 are checked.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 0. This variable will store the number of subarrays satisfying the condition.
Step 2: Iterate through the array `nums` using a `for` loop, starting from index 0 and ending at `nums.length - 3`. This ensures that there are always at least three elements remaining for a subarray of length 3.
Step 3: For each iteration, check if the condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` is met. Note that this implicitly handles integer division. In some languages like Java and C, explicit casting to a floating-point type may be needed for precise comparison to avoid potential integer truncation issues in the division.
Step 4: If the condition is true, increment the `count` variable.
Step 5: After the loop completes, return the final value of `count`.

Key Insights

Insight 1: The problem can be solved efficiently using a single loop iterating through the array. We only need to check triplets of consecutive numbers.
Insight 2: The core logic involves checking the condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` for each possible subarray of length 3. Careful consideration of integer division is needed to ensure correctness.
Insight 3: The problem has a straightforward solution; no advanced data structures or algorithms are required. The constraints on the array size (3 <= nums.length <= 100) imply that a brute-force approach is acceptable.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Cognizant.

Count Subarrays of Length Three With a Condition | LeetCode 3392

Easy

Array

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

Solution Code

Available in:

class Solution {
    public int countSubarrays(int[] nums) {
        int count = 0;
        for (int i = 0; i <= nums.length - 3; i++) {
            if (nums[i] + nums[i + 2] == nums[i + 1] * 2) {
                count++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 3392: Count Subarrays of Length Three With a Condition - Detailed explanation and solution approaches.

Topics

Array

Companies

Cognizant

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3392

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. This variable will store the number of subarrays satisfying the condition.
2
Step 2: Iterate through the array `nums` using a `for` loop, starting from index 0 and ending at `nums.length - 3`. This ensures that there are always at least three elements remaining for a subarray of length 3.
3
Step 3: For each iteration, check if the condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` is met. Note that this implicitly handles integer division. In some languages like Java and C, explicit casting to a floating-point type may be needed for precise comparison to avoid potential integer truncation issues in the division.
4
Step 4: If the condition is true, increment the `count` variable.
5
Step 5: After the loop completes, return the final value of `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 amount of work for each element. Therefore, the time complexity is linear, O(n), where n is the length of the input array. The algorithm uses only a constant amount of extra space (the `count` variable), so the space complexity is O(1).

📖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: The problem can be solved efficiently using a single loop iterating through the array. We only need to check triplets of consecutive numbers.
2Insight 2: The core logic involves checking the condition `nums[i] + nums[i + 2] == nums[i + 1] / 2` for each possible subarray of length 3. Careful consideration of integer division is needed to ensure correctness.
3Insight 3: The problem has a straightforward solution; no advanced data structures or algorithms are required. The constraints on the array size (3 <= nums.length <= 100) imply that a brute-force approach is acceptable.

📚

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: Empty or very short arrays (less than 3 elements). The code handles this implicitly by the loop condition, preventing out-of-bounds errors and returning 0 in these cases.
Edge case 2: Numbers that result in fractional values after division in `nums[i + 1] / 2`. The provided solutions implicitly handle this in differing ways. The Python solution uses floating-point comparison, which is more robust. Java and C solutions require explicit handling of potential integer truncation. Using floating-point comparisons in these languages is safer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Off-by-one errors in the loop bounds. Incorrectly setting the loop condition might lead to either missing subarrays or out-of-bounds exceptions.

✗Common mistake 2: Incorrect handling of integer division. Forgetting to handle potential truncation in integer division (particularly in Java and C solutions) leads to incorrect results for certain input values. The Python code gracefully handles this by implicit floating-point conversion, but other languages need explicit consideration of floating point types.

✅

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. This type of subarray analysis could be used to identify specific patterns in time-series data, such as identifying repeating sequences within sensor readings.

🏦Real-world scenario 2: Financial analysis. Identifying patterns in stock prices or trading volume, where the condition could represent a specific relationship between price changes over a short period.

💡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 31, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 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 Subarrays of Length Three With a Condition"?

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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