Minimum Number of Pushes to Type Word I and II | Leetcode 3014 | Leetcode 3016 | Weekly Contest

[@souravjoshi2293]

Hi MIK, I couldn't solve any Qn. Even Easy one was also very tricky to me. But I will upsolve them now. Thanks for the explanation.

[@Ramneet04]

I did this in one go, both the questions. Did the optimal one. Didn't use map in first question and used vector and did sorting in the second question. And here what I did in my first question where there is No need of map we can do only by a for loop. Int curr=1; Int ans=0; For(int i=0;i

[@imPriyansh77]

I used Max-Heap Priority Queue

[@arkojeetbera8584]

Eagerly waiting for prob4: count no. of houses at a certain distance II

[@FanIQQuiz]

Yayyyyy contest videos are here.....Thanks a lot.

[@durgeshkushwaha2534]

Really great explanation❤❤

[@souvikadhikary2429]

i took 2 hrs to solve this question. still not pass all tcs huge respect man such a awsome doubt solver

[@codestorywithMIK]

❤️❤️🙏🙏

[@Engineering.Wallah]

int ans=0; vectorv(26,0); for(auto c:word) v[c-'a']++; sort(v.begin(),v.end()); for(int i=25;i>=0;i--){ if(v[i]==0) return ans; if(25-i

[@nikhilhaspe2734]

this also works: class Solution { public int minimumPushes(String word) { int[] charCount = new int[26]; for(int i=0; i

[@pranavjawale01]

class Solution { public: int minimumPushes(string word) { unordered_map mp; for (char c : word) { mp[c]++; } if (mp.size() 8 && mp.size() < 17) { return 8 + (mp.size() - 8) * 2; } if (mp.size() > 16 && mp.size() < 25) { return 8 + 16 + (mp.size() - 16) * 3; } return 8 + 16 + 24 + (mp.size() - 24) * 4; } };

[@AryanVats603]

Did it myself easily sir!❤️

[@varanasibindu8987]

Thank you for great explanation, very helpful .

[@DevOpskagyaan]

Bhai legend ho aap 🙏🏻🔥

[@ReshuSharma-jc7yu]

MIK bhaiya pls upload weekly and bi weekly contest soln pls as the way you explain any problem give me more clarity about the topic

[@priyanshuprajapat8822]

Very Well Explained❤‍🔥

[@beenasatyendrapal6405]

🤗story ---> done

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@aizad786iqbal]

class Solution { public int minimumPushes(String word) { int n = word.length(); int ans = 0; int p=1; int x = 8; while( n >= 0){ if(n

[@anuppatankar4294]

Great video 👌

[@imPriyansh77]

Bhaiya, did today's POTD (3016) on my own...I used Max-Heap Priority Queue. Got time complexity O(n * logn) and space complexity O(26) ~ O(1)

[@codestorywithMIK]

❤️❤️❤️

[@amanpandey8031]

it was really helpful for me thanx a lot

[@thefinalfit]

Please Qn-2 also

[@priyanshumishra_IITDh]

Hello sir can you please upload solution for today's gfg potd vertex cover

[@tutuimam3381]

Thanks 👍

[@AmarjeetKumar-to9ub]

🔥

[@peterfromengland8663]

Did it on my own

[@jambajuice07]

please make video on c of biweekly contest

[@adityaraj-zm7zk]

@codestorywithMIK sir yaha pe bata dena ye kaia hua samajh nhi aaya for(char &ch: word) mp[ch - 'a']++;

[@DManikandan-cw2cw]

​ @codestorywithMIK Buddy plz upload for Biweekly Contest 122 held yesterday

[@arjitgautam365]

yes, if possible pls upload the solutions

[@codestorywithMIK]

Will try to upload soon Unfortunately couldn’t get much time ❤️🙏

[@alokbhowmik3547]

Aj ka GFG ka potd ka video bana dijia

[@salmaniproductions1104]

Thank you bhaiiya

[@dayashankarlakhotia4943]

please make video on qn.no2&no4.graph problem.

[@benoitblanc-zl8rc]

Thanks man ❤

[@souravjoshi2293]

Mik , please post Problem 2 solution.

[@amansaurabh4552]

Bhaiya plz make video on today's GFG POTD

[@FanIQQuiz]

Waiting for Qn-2 of weekly contest.

[@harshitsoni8456]

Bhai please add 2 and 4th also

[@HI-wl9el]

Doubt ::: Consider i am in a OA. Problem dp ka hai ... If my memo dp is accepted. So should i go for tabulation ???? Like that will increase my chance of selection ??? (bcoz of improved sp complexity) Or will it decrease (due to more no of submission)

[@codestorywithMIK]

No. If all test caes are accepted, run to the next qn. During face to face interview, you can discuss bottom up

[@HI-wl9el]

@@codestorywithMIK thanks, sir. But I was thinking of the scenario when all the rest questions are completed !!!

[@codestorywithMIK]

@HI-wl9el yes definitely in that case, you can try optimal approach

[@HI-wl9el]

@@codestorywithMIK thank you, sir. By the way, I love your style of explaining!

[@ashutoshpandey1639]

please make video on today's gfg potd

[@Engineering.Wallah]

int ans=0; vectorv(26,0); for(auto c:word) v[c-'a']++; sort(v.begin(),v.end()); for(int i=25;i>=0;i--){ if(v[i]==0) return ans; if(25-i

[@deeptijain1096]

please upload the solution of latest biweekly contest...........

[@rushabhlegion2560]

Yr sunday ko late kyu nhi rakhte ye contest.

[@bhuppidhamii]

I think timing is good 😊

[@rushabhlegion2560]

@@bhuppidhamii Nhi yr bro. Roz morning me office jana padta hai. Sirf sunday ek din hota hai sone ke liye.

[@bhuppidhamii]

@@rushabhlegion2560 oho

[@varunaggarwal7126]

why are we using hashmap? I did it without it class Solution { public: int minimumPushes(string word) { int ans=0; int it=0; int key=1; for(auto &ch: word){ if(it>7){ key++; it=0; } ans+=key; it++; } return ans; } };

[@adityaraj-zm7zk]

#codestorywithMIK sir iska slides daal do samhne ke liye please sir

[@tutuimam3381]

❤❤

[@dayashankarlakhotia4943]

public int minimumPushes (String word){ int[]freq=new int[26]; for(char c:word.toCharArray()){ freq[c-'a']--; } Arrays.sort(freq); int cnt=0; for(int i=0;i

[@Ankitkumar-fz3kc]

class Solution { public: int minimumPushes(string word) { int ans = 0; vector vec(26,0); for(int i = 0 ; i < word.length();i++){ vec[word[i]-'a']++; } sort(vec.rbegin(),vec.rend()); int cnt = 0; for(int i = 0 ; i< 26;i++){ if(cnt < 8 && vec[i]!=0) ans += vec[i]; else if(cnt>=8 && vec[i]!=0){ ans += ((cnt/8)+1)*vec[i]; } else{ break; } cnt++; } return ans; } }; sir this is my solution is it optimised or i can do better

[@codestorywithMIK]

This looks good Ankit. Great job.

[@faraz9641]

Merse ek b ni bna 🥲🥲

[@souravjoshi2293]

Same bhai 😥

[@rushabhlegion2560]

1st aur 3rd easy tha bro. Lekin 2nd and 4th is difficult