Kth Distinct String in an Array - Leetcode 2053 - Python

[@gmh14]

lol, your thumbnails are always interesting. you added a guitar cause the problem has "string" in its name xD

[@chrischika7026]

In python dictionaries are Ordered after 3.7

[@warrendragnir2363]

They are.

[@31redorange08]

That seems wasteful.

[@yohelln7644]

I just started doing leetcode recently as a beginner. Today I managed to do this problem at the first try and not in the worst way! I'm pretty happy right now :)

[@jeffhappens1]

Yay, I thought of and coded up the better approach for the first time. Thank you again for making these videos.

[@IntelliStar_]

FYI, dictionaries have a .get() method which takes the default value as a second parameter. So you can use it like count[s] = count.get(s, 0) + 1, if you don't want to use if statement.

[@WhosShamouz]

Yey ;) Happy for teh same solution :D class Solution: def kthDistinct(self, arr: List[str], k: int) -> str: mapping = Counter(arr) for key, value in mapping.items(): if value == 1: k -= 1 if k == 0: return key return ""

[@alexanderp7521]

Life is easier with Counter )

[@pastori2672]

i remember when i couldn't solve a hashmap problem in the google dsa course thingy and would always refer to you and be so confused how you knew it's a hashmap problem :) good times

[@Amy-601]

LibkedHashMap in Java would keep/ retain the initial order and act like a map. But python 🐍 is easier/ better. I like this one better.

[@michaelroditis1952]

you can iterate on the dictionary because in python the order is preserved

[@raphaelboakye5545]

For some reason, using two hashsets has a slight better performance than using a hashmap in Java. Can you explain why that?

[@hardiksrivastava9174]

how would you do this using 2 hashsets ?

[@blackbeard3449]

​@@hardiksrivastava9174create two hashsets, one to keep track of distinct elements and another to keep track of duplicate solutions, the second method in the editorial does this

[@leeroymlg4692]

I wouldn't trust leetcode metrics for comparing performance. Run the same code multiple times and you'll get different speeds

[@raphaelboakye5545]

@@hardiksrivastava9174 public String kthDistinct(String[] arr, int k) { Set distinct = new HashSet(); Set duplicates = new HashSet(); for(String word : arr){ if(!distinct.add(word)) duplicates.add(word); } for(String word : arr){ if(duplicates.add(word)) k--; if(k == 0) return word; } return ""; }

[@utkarshdewan8736]

If you are going by the leetcode runtime then better not because they are very random

[@juanmacias5922]

I guess dicts do keep the order the element was inserted? LOL class Solution: def kthDistinct(self, arr: List[str], k: int) -> str: count = Counter(arr) for char, cnt in count.items(): if cnt == 1: k -= 1 if k == 0: return char return ""

[@abhimanyuambastha2595]

Dicts keep the order in which the elements are entered. Dont assume it to be sorted, will cause mistakes. Use SortedContainer when needs to be sorted

[@juanmacias5922]

@@abhimanyuambastha2595 oh yeah, I guess I should have stated insertion order! xD C++ does have a sorted map, and an unordered_map though :D

[@chrischika7026]

This is True.

[@abhimanyuambastha2595]

@@juanmacias5922 Yes C has that and for python we have sortedcontainer.SortedDict and sortedcontainer.SortedList and I believe java has TreeMap

[@hrithikbandaru6462]

return [i for i in arr if Counter(arr)[i] == 1][k-1] if len([i for i in arr if Counter(arr)[i] == 1]) >= k else ""

[@dion9795]

why does this even need 2 solutions

[@utkarshdewan8736]

Useful in an interview. Starting with brute force and then moving to more optimized one. Although this question is very simple that for me as well only the second one came to mind

[@MP-ny3ep]

You have to understand that there are people who are just starting out. This approach is way better than simply explaining the optimized solution.

[@chrischika7026]

what do you mean ? the more the merrier.