LeetCode Remove All Adjacent Duplicates In String Solution Explained - Java

[@ravishraj664]

short concise and easy to understand. Your way of teaching is perfect. Please keep it up

[@smithshelke2036]

O(1) space solution ... Loop until arr(i) ==arr(i+1) then place 2 pointers a and b at I+2 and i-1 respectively ,and check if elements are equal at those indices, if yes keep expanding the pointers a and b until the elements are not equal then delete.repeat everything till we reach the end of string

[@deepakvm1475]

nice

[@yusenpeng]

Originally, I used two pointers instead of stack and it turned out to be really inefficient compared to stack. I really appreciate your stack explanation! :)

[@koikoi91]

I would like to see your solution if you dont mind please

[@krishnavar7219]

the the way you programmed this question it is genius...

[@strider_93]

How do you handle when char is repeated odd no of times like if string is abbbac

[@adhishmalviya2408]

"abac" should be the output for this I know what you're thinking output should have been "c" according to you but this problem is diffenret

[@dharmaputra7394]

Request tutorial graph like a MST, djikstra shortest path

[@044_karan9]

what was that noise in the starting of the video !!

[@sadmanabedin1203]

I thought the sound came from my side, your comment just assured me that I am safe here 😂

[@theyashbansal]

That clap at the beginning blew my ear off 😅

[@canobiggs]

You're my hero man!

[@biswaranjanbarik4052]

U r doing a very very great job thanx that I have found u in youtube. I love your videos.. Love from India..❤💌

[@dalindu7572]

why using a char[] as stack instead of using STACK ADT?

[@simonkaranja3811]

wondering too!! I did this easily using a Stack class Solution { public String removeDuplicates(String s) { Stack stack = new Stack(); StringBuilder sb = new StringBuilder(); for(int i=0;i

[@karanjindal1860]

"aaaaaaaa" this doesn't work

[@CodeSuccessChronicle]

Brother you didn’t write a 0 assignment in case there’s a duplicate how can stack change then ....

[@unknownman1]

7:25 how does string from 0,2 returns c,a. Instead of 0,2 it should be 0,1?

[@gaurav91pandey]

0 is inclusive, 2 is exclusive

[@theweekday8158]

@@gaurav91pandey actually the 2 in (stack, 0, 2 ) is the offset and not the exclusive index. 0 is the starting index and 2 here tells about the length from index 0 till which it will copy from stack into the new string.

[@canobiggs]

Such good stuff in every video

[@guyfromgalaxy]

Can do with inplace string manipulation , without using extra memory (credit to you as I learnt this trick from your previous video, so thank you! :) )

[@koikoi91]

I would like to see your solution if you dont mind please

[@ketansharma6955]

I have never used stack with indices, can someone share me some problems, which use the same. Thanks.

[@samvid74]

It does not work with aaab result is ab should be b

[@yassinesalimi2155]

very helpful. Thanks a lot !

[@satvikshukla596]

Very very helpful. Thank you!!

[@KoushikChowdhury2016]

string removeDuplicates(string s) { int i = 0, n = s.length(); for (int j = 0; j < n; ++j, ++i) { s[i] = s[j]; if (i > 0 && s[i - 1] == s[i]) // count = 2 i -= 2; } return s.substr(0, i); } This has SC: O(1)

[@abhijeetmishra4300]

wrong solution bro

[@paraskumar693]

my code is showing TLE for 9000+ characters

[@simrankak7045]

You are Amazing dude thankyou

[@sophiezhang6516]

i like your videos, but this solution doesn't work

[@varunsadhu9060]

It doesn't work for TC aabbccccbapq

[@mohdzebalam8706]

space is o(1)

[@chodingninjas7415]

u are the best

[@infotainment7123]

your code doesnt work for "geeksforgeek"

[@gauravagarwal6565]

*With no extra space* --------------------------------- StringBuilder sb = new StringBuilder(s); int i = 0; while(i0) i --; } else{ i++; } } return sb.toString();

[@nishantvaishla5752]

thanks bro

[@harshdeepsinghful]

Without extra space class Solution { public String removeDuplicates(String S) { //String s="abbaca"; for(int i=0;i

[@piyushmathpal4244]

why this line two times ? s=s.substring(0,i)+s.substring(i+1, s.length()); Also, Can you please confirm if the solution is working for the input "abbaca"?

[@harshitsharma4063]

acaaabssscp this doesn't work, I believe u thought of only two consecutive are adjacent!! Btw np , thanks keep posting

[@HarishAmarnath]

seriously you can figure out without o(n) spaces ? , you can iterate the string with 2 a and b pointers . stop the first pointer a when a and a-1 index is equal , then paste a+1th index item at a-1 , also make b=a+1; a=a-2;..... keep going until list ends.

[@yashmishra12]

Do you have the code for this approach?