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Can some body tell me whats wrong in this code vector<int> ans; ans.push_back(0); if(n==0){ return ans; } for(int i=1; i<= n; i++){ int count =0; for(int j=0; j<32; j++){ if((i & (1<<j))==1){ count++; } } ans.push_back(count); }return ans;

Thank you bhaiya!

class Solution { public int minOperations(String[] logs) { int level=1; for(String op : logs){ if(op.equals("./")){ continue; } else if(op.equals("../")){ if(level != 1){ level--; } } else{ level++; } } return level-1; } } a good simple problem to build confidence..

Super and easy Explaination! perfect! Thank you...

Thankyou bhaiya!

thank u sir

|| Java Solutions || // 1. With Stack class Solution { public int minOperations(String[] logs) { Stack<String> st = new Stack<String>(); for(int i=0; i<logs.length; i++){ if(logs[i].equals("../")){ if(!st.isEmpty()){ st.pop(); } }else if(logs[i].equals("./")){ continue; }else{ st.push(logs[i]); } } return st.size(); } } // 2. Without Stack class Solution { public int minOperations(String[] logs) { int steps=0; for(int i=0; i<logs.length; i++){ if(logs[i].equals("./")){ continue; }else if(logs[i].equals("../")){ steps = Math.max(0, --steps); }else{ steps++; } } return steps; } }

sir os, dbms jaise core subs ke lie guide kardo ya resources bata do kaise padhe for interviews?

for(string &log:logs) ka faeda kya hai & na lagae to kya farak padega ???

Approach - Initiate the count with zero. - Iterate through the whole logs and each time when "../" appears if count >0 then there will be decreament in count. - When "./" appears then leave the count as it is. - when "x/" appears increase the count by 1. - Finally when getting out of the loop if count<0 then return 0 else return count. class Solution { public: int minOperations(vector<string>& logs) { int n = logs.size(); int cnt=0; for(int i=0;i<n;i++){ if(logs[i]=="../"){ if(cnt>0) { cnt--; } } else if(logs[i]=="./")continue; else{ cnt++; } } if(cnt<0) return 0; return cnt; } };

500 days 🔥

500days man legendary

Mai apko kafi phle se follow kr rha and itta detail me dsa ke vids shyd hi koi bnata hai . Thanks bhaiya and waiting for 57k

your explanation is fabulous. your are so consistent, bhaiya can you tell me how to stay consistent.

We can also store the LCS while the table is in making when : if(s1[i-1] == s2[j-1]){ lcs += s1[i-1]; dp[i][j] = 1 + dp[i-1][j-1]; } Right?

class Solution { public: double averageWaitingTime(vector<vector<int>>& customers) { int n = customers.size(); double wait=0; double time = customers[0][0]; for(auto &it: customers){ if(time>=it[0]){ time+= it[1]; int x =time-it[0]; wait+=x; } else{ time =it[0]; time+= it[1]; int x =time-it[0]; wait+=x; } } return (wait/n); } };

Maximum Product of Word Lengths | LeetCode 318 pe vdo bna do sir

Thanks a lot

//vector<int> getNSL(vector<int>& arr, int n) , bhaiya aapne yaha NSL likh rakha hai or aap for loop se travel kr rhe right side //for(int i = 0; i<n; i++)

Bhai aise hi roz upload karte raho mai aapko dekh kar proper consistent leetcode chalu kar diya hai

Thanks a lot bhaiya ❤❤

You make it simple sir 🛐❤️and easy❤️

class Solution { public: int minOperations(vector<string>& logs) { stack<string> st; for(auto s : logs){ string str = s.substr(0,s.size()-1); cout<<str<<" "; if(!st.empty() && str == "..") st.pop(); else if(str == ".") continue; else if(str != "..") st.push(str); } return st.size(); } };❤❤

Bro how u uploaded this video at 7 or 6 am !! Like when did u even recorded and edited it

After watching ur trie video I did it myself thank u for everything

Stack is useful if asked to trace the actual path back (see Leetcode editorial)

Bhaiya Thanks for this video tooooo........can you make a video on Leetcode 306. Additive Number

It was easy 🎉

Anyone who wants to practise dsa like listening a story this is the guy u should follow :)

Good morning. I did it using stack 😊

Good morning bhaiya 😊

keep doing it ....<3

Good morning sir, just submitted my solution and got notification of your video ❤

God level of explanation by the DSA god MIK

i have a doubt...can i even solve this on my own in the future...i really had this doubt from my past 1.5 year coding experience...🙄🙄

You're Really helping the CP community. Big Kudos from my side!! ..... Please make this kind of Video. you motivated us to fight for our Goal

good

sir which writing software do you use??

bro when will i think like u

my approach class Solution { public double averageWaitingTime(int[][] arr) { int n=arr.length; double sum=arr[0][1]; int p=arr[0][0]+arr[0][1]; for(int i=1;i<n;i++){ if(arr[i][0]<p){ p+=arr[i][1]; sum+=(p-arr[i][0]); } else{ sum+=arr[i][1]; p=arr[i][0]+arr[i][1]; } } return sum/n; } }

I spent one hour and still ,I am here!

you r a gem sir

Respected MIK Sir, I solved this problem by myself. Thanks for building my logical skills. 🙇🙏 Here's my Java code: public double averageWaitingTime(int[][] customers) { int n = customers.length; double sum = 0; double previousDelay = 0; for (int[] customer: customers) { double customerArrivalTime = customer[0]; double cookingDuration = customer[1]; double cookingStartTime = Math.max(previousDelay, customerArrivalTime); double cookingEndTime = cookingStartTime + cookingDuration; double waitingTime = cookingEndTime - customerArrivalTime; previousDelay = cookingEndTime; sum += waitingTime; } return sum / n; }

Bhaiya yr aap kamaal ho daily morning uthta hu toh dekhta hu ki aapne leetcode ka potd solve krke daal diye toh esse khaafi motivation milti h ki ha bhaiya daale toh krna toh bnta h thnq bhaiya chizon ko etna aasan bnane ke liye

I made the exact same logic and on my own. Though this is easy one, but I am glad that the thought process matched.

Best Explaination for LFU😇

Based on FCFS algorithm which we studied in operating system

aaj ka question khud se kar liya... pehle mai itne bade question padh ke ghabra jaata tha.. your videos help .. bas ab series waali playlist start karni ... starting with recurison... my code intially failed beacause maine total time etc int mai le liya tha..kuch range ka issue tha I think class Solution { public double averageWaitingTime(int[][] customers) { int n = customers.length; double total_time = 0; int arrival_time = 0; int prepare_time = 0; double avg_time= 0; for(int[] x : customers){ arrival_time = x[0]; prepare_time = x[1]; if(arrival_time > total_time){ total_time += (arrival_time - total_time); } total_time += prepare_time; double waiting_time = total_time - arrival_time; avg_time += waiting_time; } return avg_time / n; } }

Solved this one without video because of your daily videos 🫡🧑‍💻. || Easy Java Solution || class Solution { public double averageWaitingTime(int[][] customers) { int totWaitTime = customers[0][0] + customers[0][1]; double waitSum = customers[0][1]; for(int i=1; i<customers.length; i++){ int[] customer = customers[i]; if(customer[0] <= totWaitTime){ totWaitTime += customer[1]; waitSum += (totWaitTime - customer[0]); } else{ totWaitTime = customer[0] + customer[1]; waitSum += customer[1]; } } return waitSum/customers.length; } }

Hello Could you please make video solution for 2407. Longest Increasing Subsequence II. This playlist is amazing🔥