explanation.md

Intersection of Two Arrays II

This code implements a solution to the "Intersection of Two Arrays II" problem, which is to find the intersection of two arrays and return it as an array. The intersection is defined as the elements that appear in both arrays, with their frequency taken into account.

Here's a step-by-step explanation of the code:

C++

1. We include the necessary headers: vector for dynamic arrays and unordered_map for hash maps.
2. We use using namespace std; to avoid having to write std:: before standard library entities.
3. The intersect function takes two vector<int> parameters: nums1 and nums2.
4. We create an unordered_map<int, int> called countMap to store the frequency of each element in nums1.
5. We create a vector<int> called result to store the intersection.
6. We iterate through nums1 and increment the count of each element in countMap.
7. We iterate through nums2:
   • For each element num in nums2, we check if it exists in countMap and its count is greater than 0.
   • If the condition is true, we append num to result and decrement its count in countMap.
8. Finally, we return the result vector containing the intersection.

Java

1. We import the necessary classes: ArrayList for dynamic arrays, HashMap for hash maps, and List for lists.
2. The intersect function takes two int[] parameters: nums1 and nums2.
3. We create a HashMap<Integer, Integer> called countMap to store the frequency of each element in nums1.
4. We create an ArrayList called result to store the intersection.
5. We iterate through nums1 and put each element num in countMap with its count.
6. We iterate through nums2:
   • For each element num in nums2, we check if it exists in countMap and its count is greater than 0.
   • If the condition is true, we append num to result and decrement its count in countMap.
7. Finally, we convert result to an int[] array and return it.

JavaScript

1. We define a function intersect that takes two parameters: nums1 and nums2.
2. We create a Map called countMap to store the frequency of each element in nums1.
3. We create an array called result to store the intersection.
4. We iterate through nums1 and put each element num in countMap with its count.
5. We iterate through nums2:
   • For each element num in nums2, we check if it exists in countMap and its count is greater than 0.
   • If the condition is true, we append num to result and decrement its count in countMap.
6. Finally, we return the result array containing the intersection.

Python

1. We import Counter from the collections module to use a frequency map.
2. We define a class Solution with a method intersect that takes two parameters: nums1 and nums2.
3. We create a Counter called countMap to store the frequency of each element in nums1.
4. We create an array called result to store the intersection.
5. We iterate through nums2:
   • For each element num in nums2, we check if its count in countMap is greater than 0.
   • If the condition is true, we append num to result and decrement its count in countMap.
6. Finally, we return the result array containing the intersection.

Go

1. We define a function intersect that takes two parameters: nums1 and nums2.
2. We create a map[int]int called countMap to store the frequency of each element in nums1.
3. We create a slice called result to store the intersection.
4. We iterate through nums1 and increment the count of each element in countMap.
5. We iterate through nums2:
   • For each element num in nums2, we check if its count in countMap is greater than 0.
   • If the condition is true, we append num to result and decrement its count in countMap.
6. Finally, we return the result slice containing the intersection.

The time complexity of this solution is O(m+n), where m and n are the lengths of nums1 and nums2, respectively. The space complexity is O(min(m,n)) since we use a hash map to store the frequency of elements from the smaller array.