What is the time complexity of Check If Array Pairs Are Divisible by k?

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

What data structures are used to solve Check If Array Pairs Are Divisible by k?

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

What are the key insights for solving Check If Array Pairs Are Divisible by k?

Insight 1: Two numbers a and b sum to a multiple of k if (a % k + b % k) % k == 0. This means a % k and b % k must sum to 0 or k. Insight 2: We can use a hash table or an array (since k is limited) to store the counts of remainders when each element is divided by k. This helps us efficiently check for pairs that sum to a multiple of k.

Which companies ask LeetCode 1497. Check If Array Pairs Are Divisible by k in interviews?

This problem is frequently asked at DevRev, Visa.

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Check If Array Pairs Are Divisible by k - LeetCode 1497 Solution

Check If Array Pairs Are Divisible by k - Complete Solution Guide

Check If Array Pairs Are Divisible by k is LeetCode problem 1497, 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

Given an array of integers arr of even length n and an integer k . We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k . Return true If you can find a way to do that or false otherwise . Example 1: Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5 Output: true Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10). Example 2: Input: arr = [1,2,3,4,5,6], k = 7 Output: true Explanation: Pairs are (1,6),(2,5) and(3,4). Example 3: Input: arr = [1,2,3,4,5

Detailed Explanation

The problem asks us to determine if an array of even length can be divided into pairs such that the sum of each pair is divisible by a given integer 'k'. The input is an array 'arr' of integers and an integer 'k'. The output is a boolean value: 'true' if it's possible to divide the array into pairs with sums divisible by 'k', and 'false' otherwise. The constraints include the even length of the array, the size of the array, the range of values in the array, and the range of the divisor 'k'.

Solution Approach

The solution uses a remainder-counting approach. First, calculate the remainder of each number in the array when divided by 'k'. Then, store these remainders' frequencies in an array. Finally, iterate through the remainders. For each remainder 'i', check if there exists a corresponding remainder 'k-i' such that their counts are equal. Handle the edge cases of remainder 0 and k/2 carefully (they must occur an even number of times). If all conditions are met, the array can be divided as required; otherwise, it cannot.

Step-by-Step Algorithm

Step 1: Initialize an array `remainder_counts` of size 'k' with all elements set to 0. This array will store the count of each remainder when an element of 'arr' is divided by 'k'.
Step 2: Iterate through the input array 'arr'. For each element 'num', calculate its remainder when divided by 'k' (num % k). If the remainder is negative, add 'k' to make it positive. Increment the corresponding count in the `remainder_counts` array.
Step 3: Check if the count of elements with remainder 0 is even. If not, return false (since they can only pair with themselves).
Step 4: Iterate from i = 1 to k/2 (inclusive). If 2 * i == k, it implies that remainder i is k/2. Check if `remainder_counts[i]` is even. If not, return false.
Step 5: If 2 * i != k, check if the count of elements with remainder 'i' is equal to the count of elements with remainder 'k-i'. If they are not equal, return false (since they can't form valid pairs).
Step 6: If all checks pass, return true (meaning the array can be arranged into pairs with sums divisible by k).

Key Insights

Insight 1: Two numbers a and b sum to a multiple of k if (a % k + b % k) % k == 0. This means a % k and b % k must sum to 0 or k.
Insight 2: We can use a hash table or an array (since k is limited) to store the counts of remainders when each element is divided by k. This helps us efficiently check for pairs that sum to a multiple of k.
Insight 3: Special cases to consider are when the remainder is 0 or k/2 (when k is even). In these cases, the count of elements with these remainders must be even for valid pairing.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: DevRev, Visa.

Check If Array Pairs Are Divisible by k | LeetCode 1497

Medium

Array Hash Table Counting

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

Solution Code

Available in:

class Solution {
    public boolean canArrange(int[] arr, int k) {
        int[] remainder_counts = new int[k];
        for (int num : arr) {
            int remainder = num % k;
            if (remainder < 0) {
                remainder += k;
            }
            remainder_counts[remainder]++;
        }

        if (remainder_counts[0] % 2 != 0) {
            return false;
        }

        for (int i = 1; i <= k / 2; i++) {
            if (2 * i == k) {
                if (remainder_counts[i] % 2 != 0) {
                    return false;
                }
            } else if (remainder_counts[i] != remainder_counts[k - i]) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1497: Check If Array Pairs Are Divisible by k - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

DevRevVisa

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1497

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: Initialize an array `remainder_counts` of size 'k' with all elements set to 0. This array will store the count of each remainder when an element of 'arr' is divided by 'k'.
2
Step 2: Iterate through the input array 'arr'. For each element 'num', calculate its remainder when divided by 'k' (num % k). If the remainder is negative, add 'k' to make it positive. Increment the corresponding count in the `remainder_counts` array.
3
Step 3: Check if the count of elements with remainder 0 is even. If not, return false (since they can only pair with themselves).
4
Step 4: Iterate from i = 1 to k/2 (inclusive). If 2 * i == k, it implies that remainder i is k/2. Check if `remainder_counts[i]` is even. If not, return false.
5
Step 5: If 2 * i != k, check if the count of elements with remainder 'i' is equal to the count of elements with remainder 'k-i'. If they are not equal, return false (since they can't form valid pairs).
6
Step 6: If all checks pass, return true (meaning the array can be arranged into pairs with sums divisible by k).

💡

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 dominant factor in time complexity is the iteration through the input array of size 'n'. The size of the `remainder_counts` array determines the space complexity. Other variables used are constant in size and do not depend on the input 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

Brute Force Approach

Try all possible pairings and check if the sum of each pair is divisible by k. This involves generating all combinations of pairs, which quickly becomes computationally expensive.

Time: O(n!) or O(n^2 * (n/2)!) (depending on implementation), where n is the length of the input array. Generating all combinations of pairs is exponential.Space: O(n) or O(1) (depending on implementation), potentially for storing the selected pairs during the search.

Trade-offs: Brute force is very inefficient and impractical for larger input sizes. The remainder-counting approach is significantly better in terms of time complexity.

✨

Key Insights

1Insight 1: Two numbers a and b sum to a multiple of k if (a % k + b % k) % k == 0. This means a % k and b % k must sum to 0 or k.
2Insight 2: We can use a hash table or an array (since k is limited) to store the counts of remainders when each element is divided by k. This helps us efficiently check for pairs that sum to a multiple of k.
3Insight 3: Special cases to consider are when the remainder is 0 or k/2 (when k is even). In these cases, the count of elements with these remainders must be even for valid pairing.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: When k is even, and the remainder is k/2. Elements with remainder k/2 can only be paired with elements that have remainder k/2. Thus the number of such elements must be even. The code explicitly checks for this condition.
Edge case 2: When the remainder is 0. Elements with remainder 0 can only be paired with elements that have remainder 0. Thus the number of such elements must be even. The code explicitly checks for this condition.
Edge case 3: Negative numbers in the input array. The code ensures the remainder is positive by adding 'k' to the remainder if it's negative, before using it as an index into the `remainder_counts` array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle negative numbers correctly when calculating the remainder. In many languages, the modulo operator can return a negative remainder if the dividend is negative. The code needs to adjust for this to ensure a positive remainder.

✗Common mistake 2: Failing to handle the case where k is even and there are remainders equal to k/2. These numbers can only be paired with each other, so their count must be even.

✗Common mistake 3: Not checking that the number of 0 remainders is even. These elements can only pair with each other.

✅

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: Load balancing: Distributing tasks among servers such that the load on each server is roughly the same. If we have tasks with different sizes and we want to divide them into groups that have a certain total size (divisible by k), we can use this logic.

🏦Real-world scenario 2: Resource allocation: Dividing resources into equal groups for distribution among different users. For example, dividing bandwidth into segments that sum to a multiple of a certain value.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 28, 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

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

@TechLeadJun 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 do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Check If Array Pairs Are Divisible by k"?

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

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