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

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

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

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

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

The core logic involves checking divisibility using the modulo operator (`%`). If `nums1[i] % (nums2[j] * k) == 0`, then `nums1[i]` is divisible by `nums2[j] * k`. A nested loop is the most straightforward approach to iterate through all possible pairs of indices `(i, j)` from `nums1` and `nums2` respectively.

Which companies ask LeetCode 3162. Find the Number of Good Pairs I in interviews?

This problem is frequently asked at Airbus SE.

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

Find the Number of Good Pairs I - LeetCode 3162 Solution

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

The provided solution uses a brute-force approach with nested loops. It iterates through each element of `nums1` and for each element, it iterates through each element of `nums2`. Inside the inner loop, it checks if the divisibility condition is met. If it is, a counter `count` is incremented. Finally, the function returns the value of `count`.

Find the Number of Good Pairs I - Complete Solution Guide

Find the Number of Good Pairs I is LeetCode problem 3162, 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

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 you to count the number of 'good pairs' between two integer arrays, `nums1` and `nums2`. A pair `(i, j)` is considered 'good' if the element at index `i` in `nums1` (i.e., `nums1[i]`) is perfectly divisible by the product of the element at index `j` in `nums2` (i.e., `nums2[j]`) and a given positive integer `k`. The input consists of two integer arrays and a positive integer `k`. The output is a single integer representing the total count of good pairs.

Solution Approach

The provided solution uses a brute-force approach with nested loops. It iterates through each element of `nums1` and for each element, it iterates through each element of `nums2`. Inside the inner loop, it checks if the divisibility condition is met. If it is, a counter `count` is incremented. Finally, the function returns the value of `count`.

Step-by-Step Algorithm

Initialize a counter variable `count` to 0.
Iterate through each element `nums1[i]` of the `nums1` array using a `for` loop.
For each `nums1[i]`, iterate through each element `nums2[j]` of the `nums2` array using another nested `for` loop.
Inside the inner loop, calculate `nums2[j] * k`.
Check if `nums1[i]` is divisible by `nums2[j] * k` using the modulo operator: `nums1[i] % (nums2[j] * k) == 0`.
If the condition is true (divisible), increment the `count`.
After iterating through all pairs, return the final value of `count`.

Key Insights

The core logic involves checking divisibility using the modulo operator (`%`). If `nums1[i] % (nums2[j] * k) == 0`, then `nums1[i]` is divisible by `nums2[j] * k`.
A nested loop is the most straightforward approach to iterate through all possible pairs of indices `(i, j)` from `nums1` and `nums2` respectively.
The problem constraints (small array sizes) suggest a brute-force approach is acceptable. More efficient solutions (e.g., using hash tables for larger inputs) are not necessary here.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Airbus SE.

Find the Number of Good Pairs I | LeetCode 3162

Easy

Array Hash Table

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

Solution Code

Available in:

class Solution {
    public int numberOfPairs(int[] nums1, int[] nums2, int k) {
        int count = 0;
        for (int i = 0; i < nums1.length; i++) {
            for (int j = 0; j < nums2.length; j++) {
                if (nums1[i] % (nums2[j] * k) == 0) {
                    count++;
                }
            }
        }
        return count;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3162: Find the Number of Good Pairs I - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Airbus SE

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3162

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

7 steps

1
Initialize a counter variable `count` to 0.
2
Iterate through each element `nums1[i]` of the `nums1` array using a `for` loop.
3
For each `nums1[i]`, iterate through each element `nums2[j]` of the `nums2` array using another nested `for` loop.
4
Inside the inner loop, calculate `nums2[j] * k`.
5
Check if `nums1[i]` is divisible by `nums2[j] * k` using the modulo operator: `nums1[i] % (nums2[j] * k) == 0`.
6
If the condition is true (divisible), increment the `count`.
7
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*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n*m) because the nested loops iterate through all possible pairs of elements from `nums1` (size n) and `nums2` (size m). The space complexity is O(1) because the algorithm uses a constant amount of extra space regardless of the input size. Only a counter variable `count` is used.

📖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

✨

Key Insights

1The core logic involves checking divisibility using the modulo operator (`%`). If `nums1[i] % (nums2[j] * k) == 0`, then `nums1[i]` is divisible by `nums2[j] * k`.
2A nested loop is the most straightforward approach to iterate through all possible pairs of indices `(i, j)` from `nums1` and `nums2` respectively.
3The problem constraints (small array sizes) suggest a brute-force approach is acceptable. More efficient solutions (e.g., using hash tables for larger inputs) are not necessary here.

📚

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

Empty arrays: The code handles empty arrays correctly; the outer loop will not execute, resulting in a count of 0.
Zero in nums2: If `nums2[j]` is 0, `nums2[j] * k` will be 0. The modulo operation with 0 will result in an error unless `nums1[i]` is also 0. The code correctly handles this case; `nums1[i] % 0` will cause an error if `nums1[i]` is not 0.

⚠️

Common Mistakes to Avoid

✗Incorrect divisibility check: Forgetting the modulo operator or using it incorrectly (e.g., `nums1[i] / (nums2[j] * k) == 0`) is a common mistake.

✗Off-by-one errors: Incorrectly defining the loop limits (e.g., using `<` instead of `<=` or vice versa) can lead to missing some pairs.

✅

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?

💼Data analysis: Finding relationships between two datasets where one dataset's values are multiples of another dataset's values (after scaling by a factor k).

🏦Signal processing: Identifying patterns in signals where one signal is a scaled and shifted version of another.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 11, 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 14

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

@CodeNinjaJul 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.

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 "Find the Number of Good Pairs I"?

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

🤖 Need help with this problem?