LeetCode - How Many Numbers Are Smaller Than the Current Number - Python

[@kirtikishore642]

Thankyou so much for such a great explanation

[@adithyapaib]

class Solution: def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]: s = sorted(nums) d= {} for i in range(len(s)): if s[i] not in d: d[s[i]] = i return [d[i] for i in nums]

[@Rednuz-88]

What is the space complexity of this problem?

[@Elfein]

O(n)

[@kirtibhushan]

Thanks man!

[@olabanji]

Thanks for the videos you make. They've all been helpful so far. Keep up the good work!

[@KnowledgeMavens]

You're welcome, thank you for watching!

[@rajeshseptember09]

Clear explanation ! Thanks. Is it possible to solve this is O(N) time ? Using extra space ? I mean without using any sorting to achieving this !

[@KnowledgeMavens]

Hey Rajesh, that's for watching and commenting. Great suggestion! Honestly, I didn't go further into this problem to consider a better space complexity given its small data set. However, it would be very beneficial to do so when implementing this algorithm on a larger set of data. Thanks again for the suggestion and I'll consider sharing more on this topic in the future.

[@rishigupta2342]

Thanks.Nice explanation!!!

[@KnowledgeMavens]

You're welcome, thank you for watching!