Find K-th Smallest Pair Distance - Leetcode 719 - Python

0:00 - Read the problem
3:25 - Intuition
11:14 - Dry Run
22:32 - Coding Solution

leetcode 149

#neetcode #leetcode #python

[NeetCodeIO]

why are the problems so hard lately, we're not even halfway through the month...

[gmh]

halfway through the video i forgot what the question was

[hardiksrivastava]

Don't worry the problem isn't even that hard.

The problem :

[sojwalgosavi]

I thought you won't drop the video, thank you Neetcode

[pk-zw]

The question states that we want pairs nums[i] and nums[j] where 0 <= i < j < nums.length. How are we taking care of this if we sort the nums array?

[ankitadey]

At first, I carefully watched your intuition explanation, had to replay few parts at times to understand properly, then when I understood what approach you talked about, I tried to code it myself. I did it. But the complexity looked pretty bad, still I was happy I did it. And then I watched you writing the code and saw that in the helper function, I wrote a longer while() loop than you because I used while() for the whole window movement instead of just l, so when i changed that part in my code, the complexity got so much better. Anyway, my point is.. THANK YOU SO MUCH FOR ALWAYS GUIDING ME! I don't watch the code right away unless it's about graph or trees because I am weak at those but you always help with the intuition part if I had been stuck for quite some time. Thanks a lot.

[MehmetDemir-xiyy]

Btw could you please explain, how to know if the value of mid in the binary search, actually exists as a difference of 2 elements in the array? We just assume there are set of differences from 0 to max but we may not have these?

[Sid]

This problem was really difficult for me lmao. It took 2-3 replays just to visualize what's going on. Thanks for the explanation!

[MehmetDemir-xiyy]

Actually brute force T.C is n^2logn since there are n * (n-1) /2 pairs

[Генри-юб]

At first time tried to implement myself. Make combinations of pairs and their differences put in an array. 17 / 19 passed

[lavanyam]

At 20:09 there are 10 pairs with difference <=2 ( you've missed the pair (4, 6) (5, 7) ) . Anyway it doesn't affect the answer as we do R = 2. Nonetheless your explanation is amazing!! Thanks:)

[siddharth-gandhi]

The only thing I'm confused about is what is the guarantee that m will be in the differences array. Not necessary that all values in [0, max] will be a possible difference of any pair. Eg. [1, 1, 6] has [0, 6] possible difference values but only 0 & 5 actually exist.

[guyguysir]

I am able to implement once understanding the method. But how on earth do I come up with this solution...

[bhumikagupta]

The brute force time complexity will be of order O(n^2 . logn) since there will be n * (n -1) / 2 distance pairs, and sorting them would have time complexity O(m . logm) where m = n * (n -1) / 2, which make it O(n^2 . logn).

[eamytb]

You should have started with the counting on the sorted array, and then optimize it by binary search, would be much better to understand

[randsam]

this is too hard yaar, not able to understand even with your explaination

[yuhangzhao]

I kinda like this question because it just combines multiple basics to solve it, not like some others solved with a crazy complicated specific algorithm.

[anandsrikumar]

I did an easy brute force and it passes 18/19 cases

[sankhadip_roy]

can do r=nums[-1] as it is already sorted, improves runtime

[ntpilrr]

Do you have to continue counting the number of pairs with differences <= M if the count exceeds k?