วีดีโอ

Big fan ❤️❤️

Thankyou Ma'am 😃

This is probably the best explanatiion of this pattern on internet now

Thanks so much for super clean explanation!

You are the best

sooo helpful !! was stuck on this question for a while...

well explained

Really it's a good explanation ,thank you mam !...

-Thank you so much for the great explanation

good work , your explanation was thorough and detailed , just one feedback it would have been great if you used map or some stl container also if you could shift to better IDEs or online platforms to solve these problems (eg vs code or sublime text) , thank you

2 on power n not x 2*n-1

Thank you ma'am

Instead of storing the values inside the array and then printing.A better approach could be to directly print the value of n

else if(s[i] == ' v' && i ! =0 && s[i-1] == ' I' ) result+=3; result += 4; chaiye tha naa mam

Best of all explanations of this question

Thanks ma'am :)

Thank you so much mam.

thank you so much

very helpful explanation and code. Thanks ma'am.

class Solution { public: int countPrimes(int n) { bool* isprime = new bool[n]; for (int i = 2; i < n; i++) { isprime[i] = true; } for (int i = 2; i * i < n; i++) { if (!isprime[i]) continue; for (int j = i * i; j < n; j += i) { isprime[j] = false; } } int count = 0; for (int i = 2; i < n; i++) { if (isprime[i]) count++; } return count; } };

Excellent mam

Thank you mam for providing the most simplest solution.

thnxx mam

Waheguru tenu Chad di kala

Good Explanation Mam.

English plzz

Best explanation for this question on TH-cam. No one has explained this question like you. Mam please upload some videos on java too.

It was Absolutely amazing

I have seen 3 or 4 solution but this one best solution 👍👍❤

Pattern printing mein array use karke kya faida, without array chahiye

I was thinking this way only ,the thing I was missing is to make two variables to iterate the loop thanku..

I can't express the way you explain the solution . You makes each complicated question an easy one. I really want to continue this series. And again a lot of thanks to you didi...

nice explaination di 🤗🤗 really able to grab the concept in a single attempt.

Great explaination and your voice is amazing

Thanks for helping in written method it was very good way understand the logic behind the program

Thankyou ma'am

code is not working

Mam,,, Please upload videos about more topics of competative programming. Really We need these because your way of explanation is best...

Can anybody tell me why we are not using printf function inside the while loop itself?

Ma'am is this the most optimized solution

Thank you man...

that's it what am searching for.

thanks

Awesome mam

Ma'am, it shows runtime error in Leetcode

One of my favourite teacher from college days 🙏🏻❤️

Alternative Solution without using 2D arrays /*C programing*/ #include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int min(int a,int b) { if(a<b) return a; else return b; } int main() { int n,m; scanf("%d", &n); for(int i=0;i<n;i++){ for(int j=0;j<n;j++) { m = min(i,j); printf("%d ",n-m); } for(int j=n-2;j>=0;j--) { m = min(i,j); printf("%d ",n-m); } printf(" "); } for(int i=n-2;i>=0;i--){ for(int j=0;j<n;j++) { m = min(i,j); printf("%d ",n-m); } for(int j=n-2;j>=0;j--) { m = min(i,j); printf("%d ",n-m); } printf(" "); } return 0; } /*C ++ programing*/ #include "bits/stdc++.h" using namespace std; int main() { int n; cin>>n; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) cout<<n-min(i,j)<<" "; for(int j=n-2;j>=0;j--) cout<<n-min(i,j)<<" "; cout<<endl; } for(int i=n-2;i>=0;i--) { for(int j=0;j<n;j++) cout<<n-min(i,j)<<" "; for(int j=n-2;j>=0;j--) cout<<n-min(i,j)<<" "; cout<<endl; } return 0; }

nicely explained :-) thank you so much ma'am

One of the best video on you tube for this question 💯