Remove Colored Pieces if Both Neighbors are the Same Color | Intuitive | GOOGLE | Leetcode - 2038

This is the 19th Video on our Greedy Playlist.
In this video we will try to solve a very good Greedy Problem - Remove Colored Pieces if Both Neighbors are the Same Color (Leetcode-2038).
Trust me, this will no longer be a Medium Problem.
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Problem Name : Remove Colored Pieces if Both Neighbors are the Same Color
Company Tags : GOOGLE
My solutions on Github : github.com/MAZ...
Leetcode Link : leetcode.com/p...
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My Graph Concepts Playlist : • Graph Concepts & Qns -...
My GitHub Repo for interview preparation : github.com/MAZ...
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[@rohankadam7376]

IF SOMEONE IS LOOKING FOR JS SOLUTION :- var winnerOfGame = function(colors) { var aliceCount = 0 var bobCount = 0 var n = colors.length; for(let i=1;ibobCount ? true : false; };

[@HarshitKumar-gn5wf]

regreting now, should thought a little more on this, such a easy question for medium level

[@gui-codes]

fabulous explanation.

[@nehakhatri3921]

Wow kya samjhaya hai

[@anuppatankar4294]

Great Video 😍

[@shabananoor9423]

❤❤❤❤❤

[@harshkishorani3101]

Man I love your explanations for Daily Leetcoding challenges! You really help me out. Thanks Man.

[@AnadiHirpara]

bhai comparator ki sereis chalu kro na mene apki playlist me dhundhi lekin nai mili

[@codestorywithMIK]

Sure thing Let me plan as soon as I get back to India

[@invinciblearmyvlogsshorts7911]

DSA has become fun becz of you...

[@onesaditya3689]

Sir we cannot remove the colors, we only count it. Sir how it print true when Alice>Bob pls explain it smthg.

[@avnish-wf8rv]

question looks so complicated but its so easy. JAVA int countA = 0; int countB = 0; for (int i = 1; i < colors.length() - 1; i++) { if (colors.charAt(i - 1) == 'A' && colors.charAt(i + 1) == 'A' && colors.charAt(i) == 'A') { countA++; } if (colors.charAt(i - 1) == 'B' && colors.charAt(i + 1) == 'B' && colors.charAt(i) == 'B') { countB++; } } if (countA > countB) { return true; } return false;

[@HARIOM_MISHRA]

Today i got toh onow how much i evolved myself in last 4months just watching ur videos today i tried to solve this question by myself and got the correct intuition same as u in my first attempt and did the question in just 2 mins i m too happy and thanku very much for evolving my thinking skill thanks alot

[@codestorywithMIK]

Thank you 🙏😇

[@keertilata20]

Was able to solve after the hints... bool winnerOfGame(string s) { int n=s.length(); if(n

[@Abhay14]

bhaiya maine ye && operator nhi lagaya maine aaise kiya mera kyun galat aaya bool winnerOfGame(string colors) { int alice = 0; int bob = 0; int n = colors.size(); for(int i=1; i bob){ return true; } return false; }

[@k.vabhavmishra]

Can any one tell me what is wrong here ? bool winnerOfGame(string colors) { int n=colors .length(); string str1="AAA"; string str2="BBB"; int countA=0,countB=0; while(n>3){ if(countA>countB){ return true; break; } if(countB>countA) { return false; break; } if(colors.find(str1)!=string::npos){ int i =colors.find(str1); countA+=1; colors.erase(i+1,1); } if(colors.find(str2)!=string::npos){ int j=colors.find(str2); countB=+1; colors.erase(j+1,1); } } return countA > countB ?true :false; }

[@Aaron-Barnes]

Thanks for simple explanation

[@utsavbarnwal7120]

Bro , sheet ki video bnao

[@Rishabhkumar-ug4qv]

First

[@RahulKumar-zp1ln]

Mastt bhai mastt.

[@codeandtalk6]

You made it so easy thank you

[@09avishkargaikwad71]

Another next level video from legend with outstanding explnation Hatts Off man!!! We would like to see your videos with leetcode weekly and Biweekly contest. Atleast for 3&4 questions ...

[@codestorywithMIK]

Thank you 🙏😇 Soon

[@oqant0424]

Explained it so well :)

[@TheAI-Tutor]

Bhaiiiiii!!!!!!!!! Hats off to you bro, Tumhe meri dua milegi

[@codestorywithMIK]

Thank you 🙏😇

[@sumitbaghel6203]

I am in BOB's team😂😂

[@dhairyachauhan6622]

# BRUTE ``` TIME: O(N^2) SPACE: O(N) ``` ``` INTUITION: - FIND THE PAIR MARK THE CENTER ELEMENT VISITED SO IT CANT BE USED AGAIN ``` ``` class Solution { public: vectorvisited; bool findPair(string &colors,char ch){ int n = colors.size(); for(int i = 1;i

[@nagmakhan3165]

The intuition was lit 🔥

[@mohithadiyal6083]

The best again😊

[@kartikrameshchavan6662]

Thanks Sir 🙏

[@jagadeeshp1163]

Done✅

[@ugcwithaddi]

Legend

[@codeandtalk6]

❤❤

[@HealthyOm]

❤❤❤❤

[@sayantanbandyopadhyay7803]

waiting for your video ...

[@killshot57]

Masterful 🙆🏻‍♂️

[@deepaksingh-qd7xm]

thought the same intuition but lazy to code 🙄🙄🥲🥲🥲🥲

[@Momentsofmagic28]

Loved the simplicity of the story. Just loved it. You are a magician.

[@floatingpoint7629]

thanks for the explanation. it was very helpful

[@AlishaKhan-ww3io]

Big fan of your stories 🙌. Thanks a lot for making this easy

[@MounikaEMB]

Very easy explanation👏 Thank you 😊

[@souravjoshi2293]

legit . You made it extremely simple

[@rounaq_khandelwal]

Your previous advice worked quite well for me!! I did well in prev contest held. and thanks for this video, was able to implement the logic as soon as i saw the greedy wallpaper at t=0

[@k.vabhavmishra]

what was the advice can u tell?

[@rounaq_khandelwal]

@@k.vabhavmishra regarding dp and recursion, dry run is very imp!

[@somsk56]

nice explanation 💗