it's been already uploaded

bantu bhai ko ram ram

kasam s ...sala karne betho to ye kese krenge...bas yahi chlta h

sahi kaha bhaiya😂koi qn sliding window me slove nahi ho raha he

@@anandunique1472 wahi sabse jyda aata hain 🤣

mt dekho tutorial.. khud se code likho fir socho kya glt h. Key Tip: Dry Run your code with pen and paper. At last, kch smjh na aae to tutorial dekho approach smjhne k liye bs.

@@rishujeetrai5780 mein to aise kr raha tha pehle video solution dekho then khud try karo

Though this code runs gets accepted in LC, but i think something is missing: if((right-left+1)-maxFreq > k){ //trimming the left portion hash[s[left]-'A']--; left++; } what if my maxFreq was from i part and also contributing, here i need to recalculate the max freq. I may be wrong but this is what i think.

@@beinginnit Yeah,I see that,but it neetcode channel also he explained in this way to optimize more to O(1) space like this,please have a look at it and explain

That he has been mentioning in his previous videos as well.

thanks bhaiya,now i got it.

Thanks a lot bro

Ok let's understand how much you really know the depth. give me pattern of all substring in ABC string with k=1.

@@kaushit {"A", "B", "C", "AB", "BC"} hi hoga na ?

yes we want greedy playlist bhaiyyaa

same bro plz give greedy playlist after this much needed

you're talking about his most optimal approach, i presume. In that case he didn't modify the maxfreq variable under the inner while loop because we don't need a maxfreq variable less than the current maxfreq variable regardless of the trimmed down substring we have right now. that's because, say, the current length of the substring is lesser but we still have the maxfreq value pertaining to the previous valid substring; in this case we simply would not require to have to check the length and change the consequent letters, we'd just ignore it. And to optimize it further, instead of looping through the letters till we get the valid window of (current_length - max_freq) > k, we could replace it with 'if' to check the condition only once because we can again simply ignore the additional trouble of having to loop through for each invalid checks which striver explains why in the other videos of his

@@sparksfly4421 thanks bhai

i agree

Trick is we dont want to minimize our max freq variable to get better result, ie. if we get result for max freq 3 we will continue to look for max freq greater than 3 if it exists and ignore freq less than equal 3. This will eliminate need of freq map scan and also remove while loop if we want to look for better max length.

@@nirmalgurjar8181 aaah i get it now ,thanks i got the 26 for loop part , but not the inner while loop part

Why would this work? Also once let's say maxfreq is 3 because of 'A', and then if we remove 'A' from the window how do we update maxfreq from 3 to 2 assuming no other char having freq 3?

​@@fractionofinfinity842actually we don't need to update the maxf value to the lesser value . Becz if u update it to lesser value like 2 , again it will come under > k while condition .

use if (r-l+1-maxf>k) rather than while

yes

I guess he did it initially as we was going to check again for the max freq using that 0-26 for loop its not in else everything is in if I guess

int[] arr = new int[26]; should be after first loop bcz in next iteration is should be updated to 0

S[r]-'A' will give a integer value which we will use for indexing in our hash map For example if s[r]=B So, 'B'-'A'(i.e 66-65(ASCII VALUES OF B AND A) =1) So hash[1] will increase

do it yourself man

@@lakshsinghania 😂😂