Construct String With Repeat Limit | Leetcode 2182

[@jatinpabari]

Thanks for the detailed explanation

[@techdose4u]

You are welcome! :)

[@Abhishek-e6d2c]

loved it!!

[@techdose4u]

great :)

[@kuriilo]

Nice one!

[@techdose4u]

Thanks :)

[@ujjwal_7531]

my code was very close but was giving rte, now its cool

[@techdose4u]

very good :) keep it up 👍🏼

[@AgastyaXV]

Hey bhaiya, I just had a few questions and wanted to know where could i contact you To summarise it up, i just wanted to know that how many hours and how many months would it take to properly learn dsa and development etc. This is coming from a 3rd sem student currently doing his bachelors Thank you!

[@techdose4u]

Since you posted your question it will be good to make a post or video :)

[@AgastyaXV]

@@techdose4u yep that would be perfect as theres many people like me. So the questions basically are: For placements in for example faang companies, what is necessary Eg dsa till trees, or leetcode level medium etc development projects weather that be gamedev/web dev etc and primarily, till what topic/what level is it necessary to learn, how long will that take in terms of hours and how many months how could we practice mock interviews etc basically to sum it up, a step by step roadmap, and how to potentially balance out college+self study

[@shekhar9595]

sir aaj ke question ki video nhi aayi

[@techdose4u]

aagyi :)

[@shekhar9595]

@@techdose4u Thanks Sir got it ♥. Sir just one request if you Hinglish in ur videos it will become more understandable

[@techdose4u]

it is expected for software engg to know english. So, consider this as your practice for placement:)

[@tmkocvines7348]

i was able to slve it by myself class Solution { public: string repeatLimitedString(string s, int l) { auto cmp = [](pair a, pair b) { return a.second < b.second; // Compare second element (char) }; unordered_mapm; // Priority queue with custom comparator priority_queue p(cmp); string ans; for(int i=0;ir) { string str(r, c); ans+=str; if(p.empty()==true) { break; } else { int a1=p.top().first; char c1=p.top().second; p.pop(); ans+=c1; p.push({a-r,c}); if(a1-1>0) p.push({a1-1,c1}); } } else { string str(a, c); ans+=str; } } return ans; } };

[@techdose4u]

nice