What is the time complexity of Special Array I?

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

What data structures are used to solve Special Array I?

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

What are the key insights for solving Special Array I?

Insight 1: The core logic revolves around checking the parity of adjacent elements. Using the modulo operator (%) to check for evenness (number % 2 == 0) or oddness (number % 2 != 0) is crucial. Insight 2: A simple iterative approach is sufficient to solve this problem. No complex data structures or algorithms are required.

Which companies ask LeetCode 3151. Special Array I in interviews?

This problem is frequently asked at National Payments Corporation of India.

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Special Array I - LeetCode 3151 Solution

Special Array I - Complete Solution Guide

Special Array I is LeetCode problem 3151, 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

An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd. You are given an array of integers nums . Return true if nums is a special array, otherwise, return false . Example 1: Input: nums = [1] Output: true Explanation: There is only one element. So the answer is true . Example 2: Input: nums = [2,1,4] Output: true Explanation: There is only two pairs: (2,1) and (1,4) , and both

Detailed Explanation

The problem asks you to determine if an array of integers is 'special'. An array is considered special if the parity (evenness or oddness) of consecutive pairs of elements is different. In simpler terms, for every pair of adjacent numbers, one must be even and the other must be odd. The input is an array of integers, and the output is a boolean value: `true` if the array is special, and `false` otherwise. The problem also specifies constraints on the input array's size and element values.

Solution Approach

The provided solutions use a straightforward iterative approach. They iterate through the input array, comparing the parity of each element with its adjacent element. If any pair has the same parity (both even or both odd), the function immediately returns `false`. If the loop completes without finding such a pair, it means all adjacent pairs have different parity, and the function returns `true`. The edge case of a single-element or empty array is handled by checking the array's length before the iteration.

Step-by-Step Algorithm

Step 1: Check if the array's length is less than or equal to 1. If so, return `true` because an array with zero or one element is considered special.
Step 2: Iterate through the array from index 0 to `length - 2`.
Step 3: For each element at index `i`, check if `nums[i] % 2` is equal to `nums[i + 1] % 2`. If they are equal (meaning both are even or both are odd), return `false`.
Step 4: If the loop completes without finding any adjacent elements with the same parity, return `true`.

Key Insights

Insight 1: The core logic revolves around checking the parity of adjacent elements. Using the modulo operator (%) to check for evenness (number % 2 == 0) or oddness (number % 2 != 0) is crucial.
Insight 2: A simple iterative approach is sufficient to solve this problem. No complex data structures or algorithms are required.
Insight 3: Handling edge cases such as an array with only one element or an empty array needs attention. A special array with one element is considered special.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: National Payments Corporation of India.

Special Array I | LeetCode 3151

Easy

Array

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

Solution Code

Available in:

class Solution {
    public boolean isArraySpecial(int[] nums) {
        if (nums.length <= 1) {
            return true;
        }
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] % 2 == nums[i + 1] % 2) {
                return false;
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 3151: Special Array I - Detailed explanation and solution approaches.

Topics

Array

Companies

National Payments Corporation of India

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3151

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

4 steps

1
Step 1: Check if the array's length is less than or equal to 1. If so, return `true` because an array with zero or one element is considered special.
2
Step 2: Iterate through the array from index 0 to `length - 2`.
3
Step 3: For each element at index `i`, check if `nums[i] % 2` is equal to `nums[i + 1] % 2`. If they are equal (meaning both are even or both are odd), return `false`.
4
Step 4: If the loop completes without finding any adjacent elements with the same parity, return `true`.

💡

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 O(n) because the algorithm iterates through the array once. The space complexity is O(1) because the algorithm uses only a constant amount of extra space, regardless of the input array's 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: The core logic revolves around checking the parity of adjacent elements. Using the modulo operator (%) to check for evenness (number % 2 == 0) or oddness (number % 2 != 0) is crucial.
2Insight 2: A simple iterative approach is sufficient to solve this problem. No complex data structures or algorithms are required.
3Insight 3: Handling edge cases such as an array with only one element or an empty array needs attention. A special array with one element is considered special.

📚

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: An array with only one element. The solution correctly handles this by returning `true` as a single element array is considered special.
Edge case 2: An empty array. While not explicitly mentioned in the problem description, the solutions implicitly handle this; the length check at the beginning prevents errors.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of an array with one or zero elements. This can lead to an index out-of-bounds error or an incorrect result.

✗Common mistake 2: Incorrectly checking parity. A common mistake is to use `if (nums[i] == nums[i+1])`, instead of comparing parity using the modulo operator `% 2`.

✅

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: Data validation: This algorithm could be used to validate data streams where alternating patterns of even and odd numbers are expected. For example, checking sensor data where values must alternate in some characteristic.

🏦Real-world scenario 2: Signal processing: In some signal processing applications, identifying alternating patterns of even and odd values might be relevant to signal detection or classification.

💡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: August 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

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Special Array 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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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