What is the time complexity of Find the Number of Good Pairs II?

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

What data structures are used to solve Find the Number of Good Pairs II?

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

What are the key insights for solving Find the Number of Good Pairs II?

Insight 1: The core of the problem lies in efficiently checking divisibility conditions between elements of the two arrays, considering the constant `k`. Insight 2: Directly iterating through all possible pairs (i, j) would result in a time complexity of O(n*m), which could be inefficient given the constraints. The given solutions reduce complexity by pre-calculating some values.

Which companies ask LeetCode 3164. Find the Number of Good Pairs II in interviews?

This problem is frequently asked at Airbus SE.

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Find the Number of Good Pairs II - LeetCode 3164 Solution

Q: How to solve Find the Number of Good Pairs II?

The provided solution pre-processes `nums1` to determine the count of multiples of each number present in `nums1`, after dividing them by `k` if divisible. Then, for each number in `nums2`, it adds to the answer the pre-computed number of its multiples in the processed `nums1`. This approach relies on pre-computation to avoid the direct pairwise divisibility checks.

Find the Number of Good Pairs II - Complete Solution Guide

Find the Number of Good Pairs II is LeetCode problem 3164, 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

You are given 2 integer arrays nums1 and nums2 of lengths n and m respectively. You are also given a positive integer k . A pair (i, j) is called good if nums1[i] is divisible by nums2[j] * k ( 0 <= i <= n - 1 , 0 <= j <= m - 1 ). Return the total number of good pairs. Example 1: Input: nums1 = [1,3,4], nums2 = [1,3,4], k = 1 Output: 5 Explanation: The 5 good pairs are (0, 0) , (1, 0) , (1, 1) , (2, 0) , and (2, 2) . Example 2: Input: nums1 = [1,2,4,12], nums2 = [2,4], k = 3 Output: 2 Explanatio

Detailed Explanation

The problem asks us to find the number of 'good' pairs between two integer arrays, `nums1` and `nums2`. A pair (i, j) is considered 'good' if `nums1[i]` is divisible by `nums2[j] * k`, where `k` is a given positive integer. The indices `i` and `j` must be within the bounds of their respective arrays (0 <= i < len(nums1) and 0 <= j < len(nums2)). We need to return the total count of such good pairs.

Solution Approach

The provided solution pre-processes `nums1` to determine the count of multiples of each number present in `nums1`, after dividing them by `k` if divisible. Then, for each number in `nums2`, it adds to the answer the pre-computed number of its multiples in the processed `nums1`. This approach relies on pre-computation to avoid the direct pairwise divisibility checks.

Step-by-Step Algorithm

Step 1: Initialize an array `counts` to store the count of each number after dividing by `k` (if divisible) from `nums1`. The size of the array is `MAX_VAL + 1`.
Step 2: Iterate through `nums1`. For each number, check if it's divisible by `k`. If it is, divide it by `k` and increment the corresponding count in the `counts` array.
Step 3: Initialize an array `multiples` to store, for each number `i`, the sum of counts of multiples of `i` present in the `counts` array. The size of the array is `MAX_VAL + 1`.
Step 4: Iterate from 1 to `MAX_VAL`. For each number `i`, iterate from `i` to `MAX_VAL` in increments of `i`. This inner loop iterates through multiples of `i`. For each multiple `j`, add `counts[j]` to `multiples[i]`.
Step 5: Initialize a variable `ans` to 0. Iterate through `nums2`. For each number, add the corresponding value from the `multiples` array to `ans`.
Step 6: Return `ans`.

Key Insights

Insight 1: The core of the problem lies in efficiently checking divisibility conditions between elements of the two arrays, considering the constant `k`.
Insight 2: Directly iterating through all possible pairs (i, j) would result in a time complexity of O(n*m), which could be inefficient given the constraints. The given solutions reduce complexity by pre-calculating some values.
Insight 3: Since `nums1[i]` must be divisible by `nums2[j] * k`, we can rephrase the condition as `nums1[i] / k` being divisible by `nums2[j]`. This suggests pre-processing nums1 by dividing its elements by k if they are divisible. We can then precompute how many multiples of each number exist among nums1/k.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Airbus SE.

Find the Number of Good Pairs II | LeetCode 3164

Medium

Array Hash Table

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

Solution Code

Available in:

class Solution {
    public long numberOfPairs(int[] nums1, int[] nums2, int k) {
        int MAX_VAL = 1000000;
        
        int[] counts = new int[MAX_VAL + 1];
        for (int num : nums1) {
            if (num % k == 0) {
                counts[num / k]++;
            }
        }
        
        long[] multiples = new long[MAX_VAL + 1];
        for (int i = 1; i <= MAX_VAL; i++) {
            for (int j = i; j <= MAX_VAL; j += i) {
                multiples[i] += counts[j];
            }
        }
        
        long ans = 0;
        for (int num : nums2) {
            ans += multiples[num];
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 3164: Find the Number of Good Pairs II - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Airbus SE

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3164

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 `counts` to store the count of each number after dividing by `k` (if divisible) from `nums1`. The size of the array is `MAX_VAL + 1`.
2
Step 2: Iterate through `nums1`. For each number, check if it's divisible by `k`. If it is, divide it by `k` and increment the corresponding count in the `counts` array.
3
Step 3: Initialize an array `multiples` to store, for each number `i`, the sum of counts of multiples of `i` present in the `counts` array. The size of the array is `MAX_VAL + 1`.
4
Step 4: Iterate from 1 to `MAX_VAL`. For each number `i`, iterate from `i` to `MAX_VAL` in increments of `i`. This inner loop iterates through multiples of `i`. For each multiple `j`, add `counts[j]` to `multiples[i]`.
5
Step 5: Initialize a variable `ans` to 0. Iterate through `nums2`. For each number, add the corresponding value from the `multiples` array to `ans`.
6
Step 6: Return `ans`.

💡

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*k)

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by iterating through the multiples to precompute the `multiples` array. The space complexity is linear with respect to MAX_VAL due to the storage of the `counts` and `multiples` arrays.

📖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

Iterate through all possible pairs (i, j) and check if `nums1[i]` is divisible by `nums2[j] * k`. Increment a counter for each good pair.

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

Trade-offs: This approach is simple to implement but inefficient for large input sizes due to its quadratic time complexity. Use it for smaller problem sizes where the complexity is not a concern.

✨

Key Insights

1Insight 1: The core of the problem lies in efficiently checking divisibility conditions between elements of the two arrays, considering the constant `k`.
2Insight 2: Directly iterating through all possible pairs (i, j) would result in a time complexity of O(n*m), which could be inefficient given the constraints. The given solutions reduce complexity by pre-calculating some values.
3Insight 3: Since `nums1[i]` must be divisible by `nums2[j] * k`, we can rephrase the condition as `nums1[i] / k` being divisible by `nums2[j]`. This suggests pre-processing nums1 by dividing its elements by k if they are divisible. We can then precompute how many multiples of each number exist among nums1/k.

📚

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: `nums1` or `nums2` being empty. The code handles this implicitly as the loops won't execute if the arrays are empty.
Edge case 2: `k` being 1. The code still works correctly, as `nums1[i]` is simply divided by 1. No special handling is required.
Edge case 3: Elements in `nums1` or `nums2` exceeding `MAX_VAL`. This will cause index out of bounds exception. Problem constraint states 1 <= nums1[i], nums2[j] <= 10^6 so it should not occur but good to be aware of.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering integer overflow when calculating `nums2[j] * k`. Using `long` data type in Java and C++ helps.

✗Common mistake 2: Incorrectly calculating the multiples in the `multiples` array. Ensure the inner loop iterates through only the multiples of `i` and not some other arbitrary numbers.

✅

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 network traffic where `nums1` represents packet sizes and `nums2` represents minimum bandwidth requirements, and `k` is a scaling factor. Finding good pairs could indicate efficient data transfer.

🏦Real-world scenario 2: Inventory management where `nums1` represents stock levels of different products and `nums2` represents customer order sizes, and `k` is a minimum order quantity. Finding good pairs could identify orders that can be fulfilled from existing stock.

💡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 13, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Find the Number of Good Pairs 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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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