Different Ways to Add Parentheses | Simple Story To Code | Leetcode 241 | codestorywithMIK

[@psychologyfact2320]

Sir contest ke solution bhi upload krna start krdo ager possible ho yo boht help hogi🙏🙏🙏

[@tanyasingh4023]

Always waiting for your explanation videos

[@robot3.077]

itne achhe tarike se koi kaise samjha sakta hai 💌💌

[@BombSquadHindiTipsTricks]

savior of stucked questions

[@akshaychavan5511]

After seeing this problem I definitely knew that it would be a recursive solution. However, I could not visualize the recursion for this problem which made it super tough for me. But thanks to MIK, now it's become super easy to understand.

[@aizad786iqbal]

nice, as per your suggestions earlier, we didn't need the solve functions, and I tried it worked with the given func itself only

[@priya_k98]

Very well explained..!

[@aws_handles]

Crystal clear explanation man.

[@AryanVats603]

Best explanation sir

[@priyobrotokarmakar6564]

Hello MIK first of all I want to thank you for your amazing teaching skill, the way you taught its really amazing.❤❤❤❤❤❤❤❤ and could you please tell what software you are using to teach us , it will really help me to take notes ❤❤

[@AshutoshAnand-o5l]

Great explaination mik!!!

[@Playvish]

Nice explaination

[@UmmeKNeha]

I always wait for your explanation. pls try to release the video early.

[@abhinay.k]

thank you

[@focusman8801]

Solved using i,j in solve Thank you bhaiya class Solution { List solve(String s,int start,int end) { List res = new ArrayList(); for(int i = start;i< end;i++) { char ch = s.charAt(i); if(ch == '-' || ch =='+' || ch =='*'){ List left = solve(s,start,i); List right = solve(s,i+1,end); for(int l : left) { for(int r : right) { if(ch == '+') { res.add(l + r); } else if(ch == '-') { res.add(l - r); } else{ res.add(l * r); } } } } } if(res.size() == 0) { res.add(Integer.valueOf(s.substring(start,end))); } return res; } public List diffWaysToCompute(String expression) { return solve(expression,0, expression.length()); } }

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@piyushgupta5044]

Bhot hi bhdiya bhai

[@thekindspill]

Guruji 👌🏻

[@ago7506]

daku for reason!!!!

[@Its_Shubham_Negi]

GOAT !!

[@anubhavsharma398]

bhaiya which mic you use for recording? Your voice is so clear.

[@footballcreativeeverywhere260]

love you bhai

[@KavyaGupta-cc5vp]

class Solution { public List diffWaysToCompute(String exp) { return solve(0,exp.length()-1,exp); } public List solve(int stIdx,int endIdx,String exp){ List list=new ArrayList(); for(int i=stIdx;i

[@dayashankarlakhotia4943]

class Solution { public: mapdp; vectordiffWaysToCompute(string exp){ vectorans; if(dp.find(exp)!=dp.end()){ return dp[exp]; } for(int i=0;i

[@SujalMahalaha]

inside first for loop in every time two inner for loop is running so time complexity will be o(n)*(2^n)*(n^2), is it right ?

[@sujayvikramgs8588]

when i was trying to come up with an idea the ideas i had was of using a stack of sorts and I realised ki that wont be feasible the other was about maybe storing it as expression trees and all, again was too complex lol idk fi thats right even

[@BatttttMan]

same, i too was thinking of stack, sort.... 😕

[@vBlxzeYT]

Bhaiya I dont understand how result.push_back(x + y) is pushing an integer in the array as we didnt convert it to int cuz last case will only work if the whole string doesnt contain any operator plz help

[@BatttttMan]

bro don't be sad, I will make it crystal clear for you 🙂 Basically, for any expression, the base case will be an expression without any operator , cuz we keep Splitting the expression whenever we are seeing operator so obv. at the end, it will be an expression without any operator. Since There will be No operator at the base, WE WILL NOT GO INTO THE LOOP and Return The Expression (no operand will be there) in the form INT (using stoi) to the Result Vector and It will be Returned to 'Left-Result' or 'Right - Result' Vector, Hence We Will Be Getting Vector of INT.

[@vBlxzeYT]

@@BatttttMan thx a lot bro ❤️

[@codestorywithMIK]

Thank you 😇❤️

[@anshror2583]

Bhai biweekly 139 k contest k 3 and 4 p bna do

[@rasish6749]

please do rotten oranges

[@jeehub041]

Sir vo har pairs ko check kar rahe uska bhi toh Time complexity hoga na and koi base case bhi nahi ha kyu?

[@nileshsinha7869]

I have one question. How the +, - and * is happening with string and getting added in integer vector

[@codestorywithMIK]

Notice that in the end when you only have digits (no operator), you will not enter for loop and in the end you will return the integer (stoi)

[@nileshsinha7869]

But let’s suppose in line 15 the x and y will be character right ? And they are getting stored in integer vector

[@rohitaggarwal8776]

Yeh question hard mai aayega not medium

[@rev_krakken70]

Was using dp for this.. is it even necessary?

[@prathmeshkakde3731]

as always nice explanation mik bhaiya

[@shobhitsingh8695]

I created separate array for numbers and operators Very similar to Unique BST 2 class Solution { public: vector numbers; vector op; map mp; vector solve(int l, int r){ if(l == r) { return {numbers[l]}; } if(mp.find({l , r}) != mp.end()){ return mp[{l , r}]; } vector res; for(int i=l;i

[@harshupadhyay2422]

can this be done using mcm?

[@RohitKumar-dz8dh]

Pattern to ussi k jesa lag raha

[@RishabhChatterjee-fg2gz]

Pehele question mein all possible ways keyword dekh Kar mere dimag mein backtracking aaya tha, fir koi approach nehi aaya dimag mein, to topics tag dekha woha dp tha par mujhe dp nehi aata, isiliye solve nehi hua, and options nehi samajh ayya question mein to recursion se kyese hoga ye bhi nehi hua. So solved after watching explanation and approach. How to find options from a recursion based question? Pls say bhaiya

[@omkarawasthi402]

Hello bhaiya, jo left and right ka recursion call kar rhe hai usme 'string' se 'int' me kaise le rha hai.

[@prathmeshkakde3731]

wo call jab tak chalega ki for loop se koi element add nahi hua aur jo last me check karega reslut.isEmpty() tab wo string to int convert kar lega

[@anammofazzaldurjoy7970]

need dp approach as well bro

[@yashnatholia2332]

class Solution { public: unordered_mapmp; vector diffWaysToCompute(string expression) { vectorans; int n = expression.size(); for(int i=0;i

[@cosmicthor7330]

Bhaeeya sab soch ney key baad code nhi likh pata any tips plz 😢

[@Coder_Buzz07]

Practice

[@madhurgupta9575]

Bhaiya, Chummi lelo, Kya video banayi h. 10 video dekh li kasam se, but this explanation was best....

[@udaysingh2929]

😅😅

[@sidhantmalik3288]

bhaiya ko toh m bhi dungaa bhaiya if possible plz dbms p video bnadoo meri job lg jyegi

[@varunpalsingh3822]

😂

[@aws_handles]

😂

[@Himanshhhhxu]

Control bhai control

[@the_only_one9415]

Hello sir, why u didnt used dp in this problem?

[@codestorywithMIK]

Definitely you can. Memoize it.

[@the_only_one9415]

@@codestorywithMIK ok sir, Thank you

[@BatttttMan]

Constraints in the Question were Light....

[@saarvikjaiswal3458]

Hello MIK bhai, I’m new to your channel and I saw great response from people watching you. I have always faced difficulties in Dsa like understanding the concept and problem related to it is not the tougher job but whenever I’ve tried solving on my own it just doesn’t happen to be. I’m requesting you to provide a sheet/combined playlist starting from basic concepts and their problems to slowly climbing the difficulty ladder to advance or med-hard problems. Please take this into consideration 🥺

[@jeehub041]

Ist view as always ❤