Intersection of Two Arrays II Leetcode daily challenge || Intuition + Code + Explanation

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.

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.

[aadarshlalchandani]

The explanation got me a success on Leetcode with this problem 🔥🔥 Thanks a ton!

[whyrohit]

you know what's more OP than the explanation, the kid behind her

[pranjalck]

I did not find the correct explanation for duplicate elements, thanks a lot it helped

[tusharnain]

This is the only video that helped me, Thank you!

[abhigyanmanasvi]

Question has been changed now. In the first example the output should be [2] and not [2, 2].

[Shva]

you are good given your age! keep it up.👍

[beinghappy]

If input arrays are sorted
Two pointer is an optimal approach

[MrKB_SSJ]

This was such an intuitive way to use hashmap! Great job

[zeyadalbadawy]

OH ! The Explaination is so Amazing !
Thanks, Alisha

[SanthoshaK-pxrq]

the approach is good but Im confusing how the code is working, please trace the code so that we could understand easily. Thanks for your effort for making edu contents

[adityasharma-iqxk]

can you take example in real world
this make your explanation is so effective
thanks a lot
I wish happy journey

[gloriadai]

very good intuition explanation! HashSet first come to my mind, but after trying, I find that it only prints out the unique elements which is no the desired answer

[AmanSingh-ffim]

In 2nd for loop hum bs yh bhi to kr skte the n If(m[num2[i]]){ vector.push_back(num2[i]) ;
m[num2[i]]--;
}

Mtlb itni saari conditions likhna nhi pdhta

[updatedtalk]

time complexity = nlogn
space complexity= n
am i right ?

[AayushSharma-whuy]

Didi can u pls help why if(mpp[nums2[i]]==1) in this line it is equal to 1 what does it indicate pls help🙏
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> mpp;
vector<int> ans;

// putting only unique values in map
for(auto i: nums1)if(mpp[i]==0)mpp[i]++;

// now if nums2[i] is present in the map -
// push it to vector & increment it's value in map to avoid repeatating element in nums2
for(int i = 0 ; i<nums2.size(); i++){
if(mpp[nums2[i]]==1){
ans.push_back(nums2[i]);
mpp[nums2[i]]++;
}
}
return ans;
}
};

[ashishp-ci]

Hi...thnx for the solution and wokrs like charm but sad part is...this is the only way to resolve this....no different ideas....everyone is solving based on this logic only.

[sudershansingh]

Thanks for the explanation, it helped!

[yogeshyts]

map, set se to nursary ke bache b kr lete h binary search se kro koi

[rohanmahajan]

What is the time and space complexity ?

[rohan]

Please provide java code of it, Thanks in advance.