What is the time complexity of Count Equal and Divisible Pairs in an Array?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Count Equal and Divisible Pairs in an Array?

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

What are the key insights for solving Count Equal and Divisible Pairs in an Array?

Insight 1: The problem requires nested loops to iterate through all possible pairs of indices. Insight 2: The modulo operator (%) is crucial for checking divisibility. `(i * j) % k == 0` efficiently determines if the product is divisible by k.

Which companies ask LeetCode 2176. Count Equal and Divisible Pairs in an Array in interviews?

This problem is frequently asked at zeta suite.

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Count Equal and Divisible Pairs in an Array - LeetCode 2176 Solution

Q: How to solve Count Equal and Divisible Pairs in an Array?

The provided code uses a brute-force approach. It iterates through all possible pairs of indices (i, j) where i < j using nested loops. For each pair, it checks if the two conditions (nums[i] == nums[j] and (i * j) % k == 0) are met. If both conditions are true, a counter `count` is incremented. Finally, the function returns the value of `count`.

Count Equal and Divisible Pairs in an Array - Complete Solution Guide

Count Equal and Divisible Pairs in an Array is LeetCode problem 2176, 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 a 0-indexed integer array nums of length n and an integer k , return the number of pairs (i, j) where 0 <= i < j < n , such that nums[i] == nums[j] and (i * j) is divisible by k . Example 1: Input: nums = [3,1,2,2,2,1,3], k = 2 Output: 4 Explanation: There are 4 pairs that meet all the requirements: - nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2. - nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2. - nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2. - nu

Detailed Explanation

The problem asks you to find the number of pairs of indices (i, j) in an integer array `nums` that satisfy two conditions: 1) The elements at those indices are equal (nums[i] == nums[j]), and 2) The product of the indices (i * j) is divisible by a given integer `k`. The pairs should be such that i < j (meaning we only consider pairs where the second index is greater than the first).

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible pairs of indices (i, j) where i < j using nested loops. For each pair, it checks if the two conditions (nums[i] == nums[j] and (i * j) % k == 0) are met. If both conditions are true, a counter `count` is incremented. Finally, the function returns the value of `count`.

Step-by-Step Algorithm

Step 1: Initialize a counter variable `count` to 0.
Step 2: Iterate through the array using nested loops: the outer loop iterates from i = 0 to n-1, and the inner loop iterates from j = i + 1 to n-1 (to ensure i < j).
Step 3: For each pair (i, j), check if nums[i] == nums[j] and (i * j) % k == 0.
Step 4: If both conditions are true, increment the `count`.
Step 5: After iterating through all pairs, return the final value of `count`.

Key Insights

Insight 1: The problem requires nested loops to iterate through all possible pairs of indices.
Insight 2: The modulo operator (%) is crucial for checking divisibility. `(i * j) % k == 0` efficiently determines if the product is divisible by k.
Insight 3: No significant optimizations are immediately apparent due to the small input constraints. A brute-force approach is sufficient.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: zeta suite.

Count Equal and Divisible Pairs in an Array | LeetCode 2176

Easy

Array

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countPairs(int[] nums, int k) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] == nums[j] && (i * j) % k == 0) {
                    count++;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2176: Count Equal and Divisible Pairs in an Array - Detailed explanation and solution approaches.

Topics

Array

Companies

zeta suite

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2176

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 counter variable `count` to 0.
2
Step 2: Iterate through the array using nested loops: the outer loop iterates from i = 0 to n-1, and the inner loop iterates from j = i + 1 to n-1 (to ensure i < j).
3
Step 3: For each pair (i, j), check if nums[i] == nums[j] and (i * j) % k == 0.
4
Step 4: If both conditions are true, increment the `count`.
5
Step 5: After iterating through all pairs, 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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through approximately n * (n-1) / 2 pairs, resulting in a time complexity of O(n^2). The algorithm uses a constant amount of extra space to store the counter variable `count`, resulting in O(1) space complexity.

📖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

Using a HashMap

Instead of nested loops, you could use a HashMap to store the frequency of each number. Then iterate through the HashMap. For each number, iterate through its indices and check the divisibility condition.

Time: O(n*log n) or O(n) depending on HashMap implementationSpace: O(n)

Trade-offs: This approach might be slightly faster for very large arrays, but the space complexity increases because of the HashMap. Given the constraints (n <= 100), the brute-force approach is likely more efficient in terms of both time and space.

✨

Key Insights

1Insight 1: The problem requires nested loops to iterate through all possible pairs of indices.
2Insight 2: The modulo operator (%) is crucial for checking divisibility. `(i * j) % k == 0` efficiently determines if the product is divisible by k.
3Insight 3: No significant optimizations are immediately apparent due to the small input constraints. A brute-force approach is sufficient.

📚

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 empty array. The code handles this correctly; the loops won't execute, and the function will return 0.
Edge case 2: k = 1. In this case, any pair with matching numbers will result in (i*j)%k == 0 since any integer is divisible by 1. The code handles this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both conditions (nums[i] == nums[j] and (i * j) % k == 0) before incrementing the counter.

✗Common mistake 2: Incorrectly handling the loop indices, leading to incorrect pair selection or double-counting.

✅

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: Analyzing datasets for similar data points that satisfy a specific index-related condition. For example, finding pairs of timestamps in log files that have the same event type and are separated by a multiple of a specific interval (e.g., checking for periodic events).

🏦Real-world scenario 2: Detecting patterns in network traffic or sensor data where the coincidence of events at specific points in time (represented by indices) and the events' similarity are important.

💡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: June 17, 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 14

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

@InterviewProJul 7

❓

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 Equal and Divisible Pairs in an Array"?

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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