What is the time complexity of Count Array Pairs Divisible by K?

The optimal solution has a time complexity of O(N + D^2).

What data structures are used to solve Count Array Pairs Divisible by K?

This problem uses Array, Math, Number Theory. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Array Pairs Divisible by K?

Insight 1: The key is to realize that instead of directly checking the divisibility of `nums[i] * nums[j]` by `k`, we can optimize by first finding the greatest common divisor (GCD) of each `nums[i]` with `k`. Insight 2: We can efficiently store and count the frequencies of these GCDs using a hash map (or dictionary).

Which companies ask LeetCode 2183. Count Array Pairs Divisible by K in interviews?

This problem is frequently asked at PayPal.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count Array Pairs Divisible by K - LeetCode 2183 Solution

Q: How to solve Count Array Pairs Divisible by K?

The solution first calculates the GCD of each number in the input array with `k`. It then counts the occurrences of each unique GCD using a hash map. After that, it iterates through all possible pairs of unique GCDs and checks if their product is divisible by `k`. If it is, it updates the count of valid pairs by considering the frequencies of each GCD. The important optimization here is computing and working with GCDs, which are typically smaller than the original numbers, and reduces the number of divisibility checks.

Count Array Pairs Divisible by K - Complete Solution Guide

Count Array Pairs Divisible by K is LeetCode problem 2183, a Hard 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) such that: 0 <= i < j <= n - 1 and nums[i] * nums[j] is divisible by k . Example 1: Input: nums = [1,2,3,4,5], k = 2 Output: 7 Explanation: The 7 pairs of indices whose corresponding products are divisible by 2 are (0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4). Their products are 2, 4, 6, 8, 10, 12, and 20 respectively. Other pairs such as (0, 2) and (2, 4) have products 3 and 15 re

Detailed Explanation

The problem asks us to count the number of pairs (i, j) in a given array `nums` such that 0 <= i < j <= n - 1 and the product of `nums[i]` and `nums[j]` is divisible by `k`. Essentially, we need to find all pairs of numbers in the array whose product is a multiple of `k`.

Solution Approach

The solution first calculates the GCD of each number in the input array with `k`. It then counts the occurrences of each unique GCD using a hash map. After that, it iterates through all possible pairs of unique GCDs and checks if their product is divisible by `k`. If it is, it updates the count of valid pairs by considering the frequencies of each GCD. The important optimization here is computing and working with GCDs, which are typically smaller than the original numbers, and reduces the number of divisibility checks.

Step-by-Step Algorithm

Step 1: Iterate through the input array `nums`.
Step 2: For each number `num` in `nums`, calculate the GCD of `num` and `k` using the `gcd(num, k)` function.
Step 3: Store the GCD and its frequency in a hash map `gcd_counts`. If the GCD already exists in the map, increment its frequency; otherwise, add it to the map with a frequency of 1.
Step 4: Extract the unique GCDs from the `gcd_counts` hash map.
Step 5: Initialize a variable `count` to 0, which will store the number of valid pairs.
Step 6: Iterate through the unique GCDs. Let `g1` and `g2` be the current pair of GCDs being considered. Ensure to consider pairs (i,j) where i <= j.
Step 7: Check if the product `g1 * g2` is divisible by `k` (i.e., `(g1 * g2) % k == 0`).
Step 8: If the product is divisible by `k`, consider the frequency counts in gcd_counts. If `g1` and `g2` are the same (i.e., i == j), add `freq * (freq - 1) / 2` to `count`, where `freq` is the frequency of `g1`. Otherwise, add `freq1 * freq2` to `count`, where `freq1` is the frequency of `g1` and `freq2` is the frequency of `g2`.
Step 9: Return the final `count`.

Key Insights

Insight 1: The key is to realize that instead of directly checking the divisibility of `nums[i] * nums[j]` by `k`, we can optimize by first finding the greatest common divisor (GCD) of each `nums[i]` with `k`.
Insight 2: We can efficiently store and count the frequencies of these GCDs using a hash map (or dictionary).
Insight 3: After having the frequencies of GCDs, we can iterate through the unique GCDs and check if the product of any two GCDs is divisible by `k`. Careful consideration is required to handle duplicate GCDs (when i==j).

Complexity Analysis

Time Complexity: O(N + D^2)

Space Complexity: O(D)

Topics

This problem involves: Array, Math, Number Theory.

Companies

Asked at: PayPal.

Count Array Pairs Divisible by K | LeetCode 2183

Hard

Array Math Number Theory

Time: O(N + D^2)Space: O(D)

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.ArrayList;
import java.util.Map;

class Solution {
    public long countPairs(int[] nums, int k) {
        Map<Integer, Integer> gcdCounts = new HashMap<>();
        for (int num : nums) {
            int gcdVal = gcd(num, k);
            gcdCounts.put(gcdVal, gcdCounts.getOrDefault(gcdVal, 0) + 1);
        }

        List<Integer> uniqueGcds = new ArrayList<>(gcdCounts.keySet());

        long count = 0;
        int nUniqueGcds = uniqueGcds.size();

        for (int i = 0; i < nUniqueGcds; i++) {
            int g1 = uniqueGcds.get(i);
            for (int j = i; j < nUniqueGcds; j++) {
                int g2 = uniqueGcds.get(j);

                if ((long)g1 * g2 % k == 0) {
                    if (i == j) {
                        int freq = gcdCounts.get(g1);
                        count += (long)freq * (freq - 1) / 2;
                    } else {
                        int freq1 = gcdCounts.get(g1);
                        int freq2 = gcdCounts.get(g2);
                        count += (long)freq1 * freq2;
                    }
                }
            }
        }

        return count;
    }

    private int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2183: Count Array Pairs Divisible by K - Detailed explanation and solution approaches.

Topics

Array Math Number Theory

Companies

PayPal

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2183

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

9 steps

1
Step 1: Iterate through the input array `nums`.
2
Step 2: For each number `num` in `nums`, calculate the GCD of `num` and `k` using the `gcd(num, k)` function.
3
Step 3: Store the GCD and its frequency in a hash map `gcd_counts`. If the GCD already exists in the map, increment its frequency; otherwise, add it to the map with a frequency of 1.
4
Step 4: Extract the unique GCDs from the `gcd_counts` hash map.
5
Step 5: Initialize a variable `count` to 0, which will store the number of valid pairs.
6
Step 6: Iterate through the unique GCDs. Let `g1` and `g2` be the current pair of GCDs being considered. Ensure to consider pairs (i,j) where i <= j.
7
Step 7: Check if the product `g1 * g2` is divisible by `k` (i.e., `(g1 * g2) % k == 0`).
8
Step 8: If the product is divisible by `k`, consider the frequency counts in gcd_counts. If `g1` and `g2` are the same (i.e., i == j), add `freq * (freq - 1) / 2` to `count`, where `freq` is the frequency of `g1`. Otherwise, add `freq1 * freq2` to `count`, where `freq1` is the frequency of `g1` and `freq2` is the frequency of `g2`.
9
Step 9: Return the final `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 + D^2)

💾

Space Complexity

O(D)

📝

Why?

The time complexity is dominated by the initial loop to calculate GCDs and the nested loops to check for divisibility. The space complexity depends on the number of distinct GCDs encountered, which in the worst case could be at most `k` (though typically much smaller). The C implementation uses fixed size arrays, so its space complexity is technically O(1), but with a large constant factor dependent on the maximum value of k. For the Python, Java and C++ version the number of unique GCD values is related to the distinct factors of K. Thus the complexity is O(D).

📖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

Iterate through all possible pairs (i, j) in the array and check if `nums[i] * nums[j]` is divisible by `k`.

Time: O(N^2), where N is the length of the `nums` array.Space: O(1)

Trade-offs: This approach is simple to understand but inefficient for large input arrays. The main solution is much better because its time complexity depends on the number of distinct GCDs which will be much less than N in most cases.

✨

Key Insights

1Insight 1: The key is to realize that instead of directly checking the divisibility of `nums[i] * nums[j]` by `k`, we can optimize by first finding the greatest common divisor (GCD) of each `nums[i]` with `k`.
2Insight 2: We can efficiently store and count the frequencies of these GCDs using a hash map (or dictionary).
3Insight 3: After having the frequencies of GCDs, we can iterate through the unique GCDs and check if the product of any two GCDs is divisible by `k`. Careful consideration is required to handle duplicate GCDs (when i==j).

📚

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: When the input array is empty. The code should correctly return 0 in this case, as there are no pairs.
Edge case 2: When all numbers in the array are divisible by `k`. In this case, every pair should be counted. The code correctly handles this scenario by counting all combinations of pairs with frequencies.
Edge case 3: When no numbers in the array are divisible by `k`. In this case, no pairs should be counted. The code correctly returns 0.
Edge case 4: Large values of `nums[i]` and `k` where the product might overflow. The Java and C++ code explicitly casts to `long` to prevent integer overflow during the multiplication of `g1` and `g2` and subsequent calculations involving frequencies. Python handles large integers automatically.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where i == j correctly. The combination formula freq * (freq - 1) / 2 must be used to avoid overcounting.

✗Common mistake 2: Integer overflow when calculating the product of `g1` and `g2` or when multiplying frequencies. This is especially important in Java, C, and C++ where you need to explicitly cast to `long`.

✗Common mistake 3: Incorrectly calculating the GCD. Ensure that the `gcd` function is implemented correctly using the Euclidean algorithm.

✅

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: This algorithm could be used in data analysis to identify pairs of events or transactions that are correlated or have a common factor (represented by `k`). For instance, identifying pairs of customers whose total spending is a multiple of a certain target value.

🏦Real-world scenario 2: In cryptography, this type of divisibility check can be used as part of a more complex key generation or encryption process. Finding number pairs with specific divisibility properties may be important in certain encryption algorithms.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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 24, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Count Array Pairs Divisible by K"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?