Leetcode 315(Hard) Count of Smaller Numbers After Self: Simple C++ Solution (Using Merge Sort)

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In this video, I provide a simple solution with an explanation to the Leetcode 315: Count of Smaller Numbers After Self in the hard category. The problem is trivial to solve with time complexity of O(N^2). This solution is based on merge sort, is supplemented with explanation and visualizations, bringing the complexity down to O(Nlog(N)).

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This is amazing. So glad to have stumbled across your video.
You can make this slightly faster by making your temp a vector of pairs, instead of a vector of vectors.
For those that do not know, traversing through a vector is slower than it is traversing through an array and in turn a pair.

satyakiranmayibhamidimarri

Best video for this question!! Keep making more videos. You will definity get more subscribers and views

ujjvalsharma

You can change the mouse settings for left handed too... world is made for both handed people.. 😀

PankajKumarThapa

Thanks for the well explained video man.

papayasprout

Thank you for the excellent video!! By the way, what is the drawing tool you are using?

hshinver

Good video 🤩. Don't give up, Please make more videos your subscribers will 📈

logicalhuman

I don't know c++ so I couldn't catch syntax errors. But I was able to find the l-r when you were coding.

a.yashwanth

For the exact same code, why i am getting TLE?

Prashantkumar-pnqq

Here is 8 line code with O(n.log(n)) complexity

Vector<int> countSmaller(vector<int>& nums) {
vector<int>v;
int n=nums.size();
vector<int>res(n);
if(n==0)
return res;
res[n-1]=0;
v.push_back(nums[n-1]);
for(int i=n-2;i>=0;i--)
{
v.insert(lower_bound(v.begin(), v.end(), nums[i]), nums[i]);
res[i]=lower_bound(v.begin(), v.end(), nums[i])-v.begin();
}

return res;
}

ayushgarg

the world is made up for right handed people.. hahaa.

SreekantShenoy

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