#code

Back in the form🔥🔥

thanks

Sir❤❤❤

Waited for you sir like crazy 😊

français formation python, ---------------------------merci

Getting TLE on this

Best teacher ever❤

Best teacher🎉

Use Dry RUn

it'Z Easy After WAtching ur video Keeps it up.

Pintu bhaiya me tiya

superb

nice

Where's the sound

Sir starting se hindi me batao jyada sahi se hoga

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Very nice 🥰❤

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❤

❤

🙏🙏🙏🙏🙏🕉️🕉️🕉️

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Very nice 🥰 bhabhishya khandelwal ❤❤❤

Very nice 🥰

Very nice 🥰 bro

Nice

Bro I am not good at maths so can I learn dsa and actually become good at it

Nice video ❤

Wahhhh chhore ❤❤❤❤

aaj ka potd 😂majak bana diya . abhi tk example testcase galat hai

I haven't tried for a two pointer approach I think if I could figure out one, it will be solvable in O(N) TC instead of O(N log N)

Code C++: Time Complexity : O(N3) int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) { vector<vector<int>>dist(n,vector<int>(n,INT_MAX)); for(int i=0;i<n;i++) dist[i][i]=0; for(vector<int>v:edges) { dist[v[0]][v[1]] = v[2]; dist[v[1]][v[0]] = v[2]; } for(int k=0;k<n;k++) //pivot { for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(dist[i][k]!=INT_MAX && dist[k][j]!=INT_MAX &&dist[i][k]+dist[k][j]<dist[i][j]) dist[i][j]= dist[i][k]+dist[k][j]; } } } int citycount; int ans; for(int i=0;i<n;i++) { int count = 0; for(int j=0;j<n;j++) { if(i!=j && dist[i][j]<= distanceThreshold) count++; } if(count<=citycount) { ans = i; citycount = count; } } return ans; }

Keep going bro👍👍👍

😄 P r o m o S M

Great bro 👍 but here can we use moore voting algorithm which time complexity O(n) and space complexity O(1) ?

For your thumbnail- you need a girl to solve those problems 😂

how does this work? 10:40

King sir is back

King sir is back

King sir is back

Love from Hyderabad ❤

Thankyou sir ❤

Sound is not getting properly

#include<iostream> using namespace std; void printArray(int arr[],int size){ cout<<"Printing the array"<<endl; for(int i=0;i<size;i++){ cout<<arr[i]<<" "; } cout<<endl; } int getMax(int arr[],int n){ int maxi=INT_MIN; for(int i=0;i<n;i++){ if(arr[i]>maxi){ maxi=arr[i]; } } return maxi; } int getMin(int arr[],int n){ int mini=INT_MAX; for(int i=0;i<n;i++){ if(arr[i]<mini){ mini=arr[i]; } } return mini; } bool search(int arr[],int n,int key){ for(int i=0;i<n;i++){ if(arr[i]==key){ return 1; } } return 0; } int main(){ int arr[15]; cout<<"value at 14 index"<<arr[14]<<" "<<endl; cout<<"value at 7 index"<<arr[7]<<" "<<endl; int arr1[3]={5,7,6}; cout<<"value at 0th index"<<arr1[0]<<" "<<endl; cout<<"value at 0th index"<<arr1[1]<<" "<<endl; int arr2[9]={4,5}; cout<<"value at 7th index"<<arr2[1]<<endl; // for(int i=0;i<10;i++){ // cout<<arr2[i]<<" "; // } // cout<<endl; printArray(arr2,10); int arr3[10]={0}; // for(int i=0;i<10;i++){ // cout<<arr3[i]<<" "; // } // cout<<endl; printArray(arr3,10); int arr4[10]={1}; // for(int i=0;i<10;i++){ // cout<<arr4[i]<<" "; // } // cout<<endl; printArray(arr4,10); // int size; // cin>>size; // int arr5[size]; // for(int i=0;i<size;i++){ // cin>>arr5[i]; // } // printArray(arr5,size); // int sizeOfarr2=sizeof(arr2)/sizeof(int); // cout<<"size of arr2"<<sizeOfarr2<<" "<< endl; // char ch[5]={'p','q','r','s','t'}; // cout<<ch[3]<<endl; //maxmin // int size; // cin>>size; // int ram[100]; // for(int i=0;i<size;i++){ // cin>>ram[i]; // } // cout<<"maximum element is"<<getMax(ram,size)<<endl; // cout<<"minimum element is"<<getMin(ram,size)<<endl; //linear search int dost[10]={-1,2,3,-4,6,8,-9,0,1,-5}; int element; cin>>element; bool found=search(dost,10,element); if(found){ cout<<"Element is present in array"<<endl; } else{ cout<<"Element is absent in array"<<endl; } }

nice explain

Long time....

Nice explanation as well as algorithmic approach

Great Video Bro !

This one takes O(n^2) time. A faster solution is: #include <iostream> #include <string> using namespace std; class Solution { public: string reverseWords(string str) { int n = str.size(); string result{""}; string temp{""}; for (int i = 0; i < n; i++) { if (str[i] != ' ') { temp = str[i] + temp; } else { result = result + temp + " "; temp = ""; } } return result + temp + "\0"; } }; int main() { Solution sol; cout << sol.reverseWords("This is a sentence "); return 0; } This one does it in linear time