What is the time complexity of Intersection of Two Arrays II?

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

What data structures are used to solve Intersection of Two Arrays II?

This problem uses Array, Hash Table, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Intersection of Two Arrays II?

Insight 1: We need to count the frequency of each number in at least one of the arrays to determine how many times it appears in the intersection. Insight 2: A hash table (or dictionary) is an efficient data structure for storing the frequency counts of the numbers.

Which companies ask LeetCode 350. Intersection of Two Arrays II in interviews?

This problem is frequently asked at Yandex.

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Intersection of Two Arrays II - LeetCode 350 Solution

Q: How to solve Intersection of Two Arrays II?

The solution uses a hash table (dictionary in Python) to store the frequency of each element in the first array (`nums1`). Then, it iterates through the second array (`nums2`). For each element in `nums2`, it checks if the element exists in the hash table and if its count is greater than 0. If both conditions are true, it appends the element to the result array and decrements its count in the hash table. This ensures that the number appears in the result as many times as it appears in both arrays. Finally, the result array is returned.

Q: What is the space complexity of Intersection of Two Arrays II?

The space complexity is O(min(m,n)).

Intersection of Two Arrays II - Complete Solution Guide

Intersection of Two Arrays II is LeetCode problem 350, 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 two integer arrays nums1 and nums2 , return an array of their intersection . Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order . Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9] Explanation: [9,4] is also accepted. Constraints: 1 <= nums1.length, nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 1000 Follow up: What if the given array i

Detailed Explanation

The problem asks us to find the intersection of two integer arrays, `nums1` and `nums2`. The intersection should include each number that appears in *both* arrays, and the number of times it appears in the result must match the *minimum* number of times it appears in either `nums1` or `nums2`. The order of the elements in the result array doesn't matter. The constraints specify that the array lengths are at most 1000, and the values in the arrays are between 0 and 1000.

Solution Approach

The solution uses a hash table (dictionary in Python) to store the frequency of each element in the first array (`nums1`). Then, it iterates through the second array (`nums2`). For each element in `nums2`, it checks if the element exists in the hash table and if its count is greater than 0. If both conditions are true, it appends the element to the result array and decrements its count in the hash table. This ensures that the number appears in the result as many times as it appears in both arrays. Finally, the result array is returned.

Step-by-Step Algorithm

Step 1: Create an empty hash table (dictionary) called `count`.
Step 2: Iterate through `nums1`. For each element `num` in `nums1`, increment its count in the `count` hash table. If `num` is not already in `count`, initialize its count to 1.
Step 3: Create an empty list called `result` to store the intersection elements.
Step 4: Iterate through `nums2`. For each element `num` in `nums2`:
Step 5: Check if `num` exists as a key in the `count` hash table AND if `count[num]` is greater than 0.
Step 6: If both conditions in Step 5 are true:
Step 7: Append `num` to the `result` list.
Step 8: Decrement `count[num]` by 1.
Step 9: After iterating through all elements in `nums2`, return the `result` list.

Key Insights

Insight 1: We need to count the frequency of each number in at least one of the arrays to determine how many times it appears in the intersection.
Insight 2: A hash table (or dictionary) is an efficient data structure for storing the frequency counts of the numbers.
Insight 3: We need to iterate through the second array and check if each number is present in the hash table and if its count is greater than zero. If so, we add it to the result and decrement its count in the hash table.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(min(m,n))

Topics

This problem involves: Array, Hash Table, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Yandex.

Intersection of Two Arrays II | LeetCode 350

Easy

Array Hash Table Two Pointers Binary Search Sorting

Time: O(m+n)Space: O(min(m,n))

Solution Code

Available in:

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums1) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        List<Integer> result = new ArrayList<>();
        for (int num : nums2) {
            if (map.containsKey(num) && map.get(num) > 0) {
                result.add(num);
                map.put(num, map.get(num) - 1);
            }
        }
        int[] res = new int[result.size()];
        for (int i = 0; i < result.size(); i++) {
            res[i] = result.get(i);
        }
        return res;
    }
}

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Problem Info

Description

Problem 350: Intersection of Two Arrays II - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers Binary Search Sorting

Companies

Yandex

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:350

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: Create an empty hash table (dictionary) called `count`.
2
Step 2: Iterate through `nums1`. For each element `num` in `nums1`, increment its count in the `count` hash table. If `num` is not already in `count`, initialize its count to 1.
3
Step 3: Create an empty list called `result` to store the intersection elements.
4
Step 4: Iterate through `nums2`. For each element `num` in `nums2`:
5
Step 5: Check if `num` exists as a key in the `count` hash table AND if `count[num]` is greater than 0.
6
Step 6: If both conditions in Step 5 are true:
7
Step 7: Append `num` to the `result` list.
8
Step 8: Decrement `count[num]` by 1.
9
Step 9: After iterating through all elements in `nums2`, return the `result` list.

💡

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(m+n)

💾

Space Complexity

O(min(m,n))

📝

Why?

The time complexity is O(m+n) because we iterate through `nums1` once (O(m)) to populate the hash table and then iterate through `nums2` once (O(n)) to find the intersection. The space complexity is O(min(m,n)) because the hash table will store at most the distinct elements of the shorter array. In the worst case where there are no common elements, the hash table stores at most `min(m, n)` distinct numbers.

📖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

Sorting and Two Pointers

Sort both arrays and then use two pointers to find common elements. Increment the smaller pointer until a match is found or the pointer reaches the end of the array.

Time: O(m log m + n log n), where m and n are the lengths of nums1 and nums2, respectively (due to sorting). If the arrays are already sorted then it's O(m+n)Space: O(1) if the sorting is done in-place. O(m) or O(n) depending on the sorting algorithm, if not in place.

Trade-offs: This approach is better if the arrays are already sorted or if in-place sorting is possible and space complexity is a major concern. If the arrays need to be sorted first, the hash table approach is often faster due to the lower constant factors, but sorting and two pointers is good with sorted data.

✨

Key Insights

1Insight 1: We need to count the frequency of each number in at least one of the arrays to determine how many times it appears in the intersection.
2Insight 2: A hash table (or dictionary) is an efficient data structure for storing the frequency counts of the numbers.
3Insight 3: We need to iterate through the second array and check if each number is present in the hash table and if its count is greater than zero. If so, we add it to the result and decrement its count in the hash table.

📚

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!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input arrays. If either `nums1` or `nums2` is empty, the intersection is an empty array. The code handles this correctly because the loops simply don't execute, resulting in an empty `result` array.
Edge case 2: Duplicate numbers in the input arrays. The solution correctly handles this by storing the frequency of each number in the hash table. The intersection will contain the numbers with their respective multiplicities based on the minimum count of each number in the two input arrays.
Edge case 3: One array contains a large number of unique elements and the other contains a single element. The solution still operates correctly because the hashmap will grow based on the elements in the first array and iterate through the second array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate numbers correctly. Forgetting to track the frequency of each number can lead to incorrect results if the number appears multiple times in either array.

✗Common mistake 2: Returning unique elements instead of the intersection with correct counts. The problem explicitly states that the result should include elements as many times as they appear in both arrays, so simply finding the unique common elements is not sufficient.

✗Common mistake 3: Modifying the arrays in-place instead of creating a new array to store the intersection. This can lead to unexpected behavior if the input arrays are needed later.

✅

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: Database queries. Finding the intersection of two sets of results from different database tables.

🏦Real-world scenario 2: Recommendation systems. Finding items that both a user has previously interacted with and items that are currently trending, to suggest relevant content.

🎮Real-world scenario 3: Data analysis. Identifying common customers between two different marketing campaigns to analyze their behavior and optimize future campaigns.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: November 18, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

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 "Intersection of Two Arrays II"?

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.

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