Kth Distinct String in an Array - Leetcode 2053 - Python

0:00 - Read the problem
0:31 - Drawing Explanation
2:31 - Coding Explanation
5:01 - Drawing Explanation 2
6:32 - Coding Explanation 2

leetcode 2053

#neetcode #leetcode #python

[gmh]

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

[yohelln]

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 :)

[jeffhappens]

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

[chrischika]

In python dictionaries are Ordered after 3.7

[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 ""

[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.

[pastori]

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-]

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

[michaelroditis]

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

[raphaelboakye]

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

[juanmacias]

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 ""

[hrithikbandaru]

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 ""