What is the time complexity of Special Array II?

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

What data structures are used to solve Special Array II?

This problem uses Array, Binary Search, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Special Array II?

Insight 1: The core challenge is efficiently checking the parity of adjacent elements within the given subarray for each query. Insight 2: Prefix sums can be used to precompute the number of parity violations (adjacent elements with the same parity) up to each index, allowing for O(1) query processing after the O(n) precomputation.

Which companies ask LeetCode 3152. Special Array II in interviews?

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

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Special Array II - LeetCode 3152 Solution

Special Array II - Complete Solution Guide

Special Array II is LeetCode problem 3152, 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

An array is considered special if every pair of its adjacent elements contains two numbers with different parity. You are given an array of integer nums and a 2D integer matrix queries , where for queries[i] = [from i , to i ] your task is to check that subarray nums[from i ..to i ] is special or not. Return an array of booleans answer such that answer[i] is true if nums[from i ..to i ] is special. Example 1: Input: nums = [3,4,1,2,6], queries = [[0,4]] Output: [false] Explanation: The subarray

Detailed Explanation

The problem asks you to determine, for a given array of integers `nums` and a list of queries `queries`, whether the subarray specified by each query is "special". A subarray is considered "special" if every pair of adjacent elements within that subarray has different parity (i.e., one is even and the other is odd). Each query `queries[i]` contains a start and end index [from_i, to_i] indicating the subarray `nums[from_i...to_i]` to be tested. The output is an array of booleans, where each element indicates whether the corresponding queried subarray is special (true) or not (false). The array indices are inclusive.

Solution Approach

The provided solution uses a prefix sum technique to efficiently determine if a subarray is special. It first calculates a prefix sum array `prefix_violations`, where each element `prefix_violations[i]` stores the number of times adjacent elements in `nums[0...i]` have the same parity. Then, for each query `[from_i, to_i]`, it subtracts `prefix_violations[from_i]` from `prefix_violations[to_i]` to obtain the number of parity violations within the subarray `nums[from_i+1...to_i]`. If this count is zero and the `from_i` and `to_i` indices are different, or the `from_i` and `to_i` are same the subarray is considered special, if `from_i` and `to_i` are different, and the violation count is zero, then the subarray `nums[from_i...to_i]` does not have any adjacent element with same parity and is therefore special.

Step-by-Step Algorithm

Step 1: Create a `prefix_violations` array of the same length as `nums`, initialized with zeros.
Step 2: Iterate through `nums` from index 1 to `n-1`. For each index `i`, calculate the number of violations up to that point. A violation occurs if `nums[i-1]` and `nums[i]` have the same parity. Update `prefix_violations[i]` as `prefix_violations[i-1] + (1 if nums[i-1]%2 == nums[i]%2 else 0)`.
Step 3: Create an `answer` array to store the boolean results for each query.
Step 4: Iterate through the `queries` array. For each query `[from_i, to_i]`, calculate the number of violations within the subarray `nums[from_i...to_i]`. If `from_i` and `to_i` are equal then the subarray only contains one element and it is special, so add 'true' to answer. Otherwise set `violations_count` to `prefix_violations[to_i] - prefix_violations[from_i]`
Step 5: If `violations_count` is 0, it means there are no adjacent element with same parity and add `true` to the `answer` array. Otherwise, add `false` to the `answer` array.
Step 6: Return the `answer` array.

Key Insights

Insight 1: The core challenge is efficiently checking the parity of adjacent elements within the given subarray for each query.
Insight 2: Prefix sums can be used to precompute the number of parity violations (adjacent elements with the same parity) up to each index, allowing for O(1) query processing after the O(n) precomputation.
Insight 3: An edge case to consider is when the 'from' and 'to' indices are equal, meaning the subarray contains only one element, and hence it is always considered special.

Complexity Analysis

Time Complexity: O(n + q)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Prefix Sum.

Companies

Asked at: National Payments Corporation of India.

Special Array II | LeetCode 3152

Medium

Array Binary Search Prefix Sum

Time: O(n + q)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean[] isArraySpecial(int[] nums, int[][] queries) {
        int n = nums.length;
        
        int[] prefixViolations = new int[n];
        for (int i = 1; i < n; i++) {
            prefixViolations[i] = prefixViolations[i-1];
            if ((nums[i-1] % 2) == (nums[i] % 2)) {
                prefixViolations[i] += 1;
            }
        }
        
        boolean[] answer = new boolean[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int fromi = queries[i][0];
            int toi = queries[i][1];
            
            if (fromi == toi) {
                answer[i] = true;
                continue;
            }
            
            int violationsCount = prefixViolations[toi] - prefixViolations[fromi];
            
            answer[i] = (violationsCount == 0);
        }
        
        return answer;
    }
}

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Problem Info

Description

Problem 3152: Special Array II - Detailed explanation and solution approaches.

Topics

Array Binary Search Prefix Sum

Companies

National Payments Corporation of India

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3152

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

6 steps

1
Step 1: Create a `prefix_violations` array of the same length as `nums`, initialized with zeros.
2
Step 2: Iterate through `nums` from index 1 to `n-1`. For each index `i`, calculate the number of violations up to that point. A violation occurs if `nums[i-1]` and `nums[i]` have the same parity. Update `prefix_violations[i]` as `prefix_violations[i-1] + (1 if nums[i-1]%2 == nums[i]%2 else 0)`.
3
Step 3: Create an `answer` array to store the boolean results for each query.
4
Step 4: Iterate through the `queries` array. For each query `[from_i, to_i]`, calculate the number of violations within the subarray `nums[from_i...to_i]`. If `from_i` and `to_i` are equal then the subarray only contains one element and it is special, so add 'true' to answer. Otherwise set `violations_count` to `prefix_violations[to_i] - prefix_violations[from_i]`
5
Step 5: If `violations_count` is 0, it means there are no adjacent element with same parity and add `true` to the `answer` array. Otherwise, add `false` to the `answer` array.
6
Step 6: Return the `answer` array.

💡

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 + q)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The prefix sum calculation iterates through `nums` once. The query processing iterates through `queries` once, with each query being processed in constant time due to the precomputed prefix sums. The space used is dominated by the size of the `prefix_violations` array, which is proportional to the size of `nums`.

📖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

For each query, iterate through the subarray `nums[from_i...to_i]` and check if adjacent elements have different parity. If any pair has the same parity, the subarray is not special.

Time: O(n*q) where n is the average length of subarray, or the length of `nums` in the worst case, and q is the number of queriesSpace: O(1)

Trade-offs: The brute force approach has lower space complexity but higher time complexity. It is only suitable for small input sizes where the number of queries and the size of the array are limited. The prefix sum approach is much more efficient for larger input sizes.

✨

Key Insights

1Insight 1: The core challenge is efficiently checking the parity of adjacent elements within the given subarray for each query.
2Insight 2: Prefix sums can be used to precompute the number of parity violations (adjacent elements with the same parity) up to each index, allowing for O(1) query processing after the O(n) precomputation.
3Insight 3: An edge case to consider is when the 'from' and 'to' indices are equal, meaning the subarray contains only one element, and hence it is always considered special.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: `from_i == to_i`. In this case, the subarray contains only one element. The problem statement suggests this case is considered 'special'. The solution correctly handles this by directly adding `true` to the `answer` array.
Edge case 2: An empty `nums` array is technically not possible according to constraints, but if it were the code should still function due to the constraints. The code would return an empty `answer` if `queries` is empty, according to constraints this can happen.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to account for the edge case where `from_i == to_i`. This can lead to incorrect results for single-element subarrays.

✗Common mistake 2: Incorrectly calculating the number of violations within a subarray when using prefix sums. Make sure to subtract `prefix_violations[from_i]` from `prefix_violations[to_i]`, and be mindful of whether the indices are inclusive or exclusive.

✅

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. Checking for consistency in data sequences, such as ensuring alternating types in a data stream.

🏦Real-world scenario 2: Signal processing. Identifying patterns or anomalies in signals where alternating high and low values are expected. The prefix-sum-based algorithm, or related dynamic programming principles are common place.

💡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: September 1, 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 do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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

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