Minimum window substring | LeetCode Hard problem | Detailed solution with code | DSA

[@JAYAPUSHPAMV]

keep making video one of the best video for sliding window

[@sharad3877]

best explaination on YT for this problem

[@anirudhm187]

Helpful , best explaination on YT ! 👏👏👏👏

[@SaadKhan-sg4wh]

clean and readable code, good work!

[@empiregaming3675]

explanation at speed of 1.75 is best to understand and save time

[@hitendusseja6615]

Hello you have initialised the vector ( the map to store frequencies) by 0, but later on in the video you said the frequency of elements which are not in 't' string would be -1. Is there a correction that I am missing sir?

[@world4coding]

You should have to be cool and think about the approach he is using and do 2-3 dry runs. He is actually explaining best on entire youtube.

[@hardcash9930]

what if string s contains AAA in the starting count will increase each time then after that count will be equal to s ka size it will update minlenth and ans will be zero plz explain request

[@cricophobiya4178]

from collections import Counter def minimum_window(str1,str2): window="a"*(len(str1)+1) left=0 count=0 hashmap=Counter(str2) for i in range(len(str1)): if str1[i] in hashmap: hashmap [str1[i]]-=1 if hashmap [str1[i]]>=0: count+=1 while count==len(str2): if len(str1[left:i+1]) < len(window): window=str1[left:i+1] if str1[left] in hashmap: hashmap [str1[left]]+=1 if hashmap [str1[left]]>0: count-=1 left+=1 return window str1='ADOBECODEBANC' str2='ABC' print(minimum_window(str1,str2)) str1 = "PRWSOERIUSFK" str2 = "OSU" print(minimum_window(str1,str2))

[@shubhamsukum]

Thanks! It helped me. ^_^

[@Om754]

thx

[@surojitmodak1683]

#bestexplanation

[@its_me_hb]

not helpful

[@world4coding]

You should have to be cool and think about the approach he is using and do 2-3 dry runs. He is actually explaining best on entire youtube.

[@alhassanecamara2095]

He also do his best to make the video sort and easy to watch!