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# Intersection of Two Arrays II LeetCode Solution

## Problem – Intersection of Two Arrays II

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`

• What if the given array is already sorted? How would you optimize your algorithm?
• What if `nums1`‘s size is small compared to `nums2`‘s size? Which algorithm is better?
• What if elements of `nums2` are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

### Intersection of Two Arrays II LeetCode Solution in Python

``````class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
if len(nums1) > len(nums2): return self.intersect(nums2, nums1)

cnt = Counter(nums1)
ans = []
for x in nums2:
if cnt[x] > 0:
ans.append(x)
cnt[x] -= 1
return ans
``````

### Intersection of Two Arrays II LeetCode Solution in Java

``````public class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
ArrayList<Integer> result = new ArrayList<Integer>();
for(int i = 0; i < nums1.length; i++)
{
if(map.containsKey(nums1[i])) map.put(nums1[i], map.get(nums1[i])+1);
else map.put(nums1[i], 1);
}

for(int i = 0; i < nums2.length; i++)
{
if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0)
{
map.put(nums2[i], map.get(nums2[i])-1);
}
}

int[] r = new int[result.size()];
for(int i = 0; i < result.size(); i++)
{
r[i] = result.get(i);
}

return r;
}
}
``````

### Intersection of Two Arrays II LeetCode Solution in C++

``````class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> dict;
vector<int> res;
for(int i = 0; i < (int)nums1.size(); i++) dict[nums1[i]]++;
for(int i = 0; i < (int)nums2.size(); i++)
if(--dict[nums2[i]] >= 0) res.push_back(nums2[i]);
return res;
}
};
``````
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