Mam your voice sounds very similar to mine

hey ayushi where are you from ?? and you are in which college ??

C++ ka code nhi daalogi to etne hi views aayenge.

C++ code ?

If we start from the starting it will have same time complexity as start from the end. Why we should not choose 1st approach?

Shouldn't it be returning i+1 because as we need the length

link

Thank you!!

00:09 - Introduction to Gray Code and its unique properties. 00:56 - Gray Code differences and problem constraints. 01:42 - Introduction to binary representation and Gray Code generation. 02:39 - Constructing Gray Code for n equals 2 through binary additions. 03:36 - Exploring Gray Code generation for n equals 3. 04:33 - Gray Code sequences differ by one bit at each transition. 05:29 - Explains logic implementation for generating Gray Code values. 06:18 - Explains Gray Code generation by iterating through values in reverse.

cool, nice explanation 😄

very insightful for me.

😄 cool solution

my column indicator changed from A B C D to 1 2 3 4. How to change it back?

Hey Ayushi, you said in one of your linkedIn posts that when you were targeting a switch one of the companies you were targeting was Visa. Visa doesn't give free food like Walmart does, is that not a cons that bothers you significantly? Cuz for some people food is indeed a big deal.

nice explanation, I really like your dry run approach before code in almost all your uploads. Someone once said "If we cannot solve a problem by ourselves, don;t expect the computer to solve it for us", sounds like simple statement, but very true.

THANKYOU SOO MUCH FOR PUTTING THAT MUCH EFFORT IN MAKING THIS SHEET , IT IS HELPING ME ALOT . HOPE TO MEET YOU ONE DAY INPERSON. YOU'RE AN INSPIRATION TO ME REALLY .

Nice method.

In leetcode this is in easy section but here you have kept in hard why?

I appreciate the solution, but unfortunately, it does not cover the scenarios where some nodes are unvisited neighbors. How will we handle that situation, and what will be the ordering of the traversal, all those possibilities are unanswered because of this specific case. The following scenario can help cover most of the scenarios: [[1,0],[0,3],[0,2],[3,2],[2,5],[4,5],[5,6],[2,4]]

If You are Small Girl and Success Had A face

Have you considered the fact that you are a girl

like my comment

working code with T=O(n) and S=O(1) int wiggleMaxLength(vector<int>& nums) { int n=nums.size(); if(n==1)return n; int i=1, prev_diff=0, cnt=1; //handling duplicate element edge case in start to calculate the prev_diff while(i<n&&prev_diff==0){ if(nums[i]-nums[i-1]!=0){prev_diff=nums[i]-nums[i-1];cnt++;} i++; } while(i<n){ if((prev_diff>0&&nums[i]-nums[i-1]<0)||(prev_diff<0&&nums[i]-nums[i-1]>0)){ cnt++;prev_diff=nums[i]-nums[i-1]; } i++; } return cnt; }

Hi, This video is so helpful. Can you please create a video on top internships for students like me? I saw this Deconstruct ₹1L internship recently, and it got me interested in skincare-related internships.

class Solution { public int myAtoi(String s) { //remove leading zeroes StringBuilder sb = new StringBuilder(s); while(sb.length()>0 && Character.isWhitespace(sb.charAt(0))){ sb.deleteCharAt(0); } if(sb.length()==0) return 0; boolean neg = false; char ch = sb.charAt(0); if(ch=='-'){ neg = true; sb.deleteCharAt(0); } if(ch=='+'){ sb.deleteCharAt(0); } StringBuilder sb2 = new StringBuilder(); for(int i=0; i<sb.length(); i++){ char ch2 = sb.charAt(i); if(ch2>='0' && ch2<='9'){ sb2.append(ch2); }else break; } if(sb2.length()==0) return 0; int res = 0; try{ res = Integer.parseInt(sb2.toString()); }catch(Exception e){ if(neg) return Integer.MIN_VALUE; else return Integer.MAX_VALUE; } if(neg){ res = -1 * res; } return res; } } Used exception handling for easy way out. lol

Why does the for loop works? How can we guarentee that the case doesn't fail till gas.size -1? like what if the answer for first question was index = 4?

nice

i just wanna say one thing to you NEVER don't give up "tum karlogi yaar"

how can these program achieve by java 8?

I initially did the tot cost thinf but what i did was , took the diff and stored in gas[i] and then I check if sum of gas array is < 0 , I did this on my own and got amazed my original soln beat 15 % , but after video I got clicked that Oh, I do not need to check the whole circle if the previous indices are not ans , then upcoming must . Nice explanation . /*I have solved this question in O(n) but still it beats 15% , so I referred one yt video and to my surprise I brilliantly handled the edge case just that mine main loop needed a bit change . I was actually checking till cirucular loop if the ans is getting found , but the catch is that , for ex : firstly , we check whether the sum of diff <0 ; if yes then the ans is not at all possible . next , if its possible and index 0,1,2 are not starting points 4 and 5 must be there , so increment the start accordingly and whenver the case fails just increment the start . so , need not to check from start again . */ class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int sum = 0 ; int n = gas.length ; // int i=0 ; int start = 0 ; // int cnt = 0; int tot = 0; //firstly storing the difff . for(int k=0 ; k<gas.length ; k++){ gas[k] = gas[k] - cost[k] ; tot += gas[k] ; } //if tot < 0 , hence the circular track not completed . //OG statement , it helped me to solve the large testcases . if(tot < 0) return -1 ; // while(cnt != n && i < n ){ //cnt == n , tells the circle completed . && i == n tells the rejection is = to n . // if(sum + gas[start] >= 0){ // sum += gas[start] ; //storing sum . // cnt++ ; // } // else{ // cnt = 0 ; //reset sum and cnt . // sum = 0 ; // i++ ; // } // start = (start+1) % n ; //this avoids mess and cleanly handles the error . // } // if(cnt == n) return start ; //it means all stations visited and returned to same station(index) . for (int j = 0; j < n; j++) { sum += gas[j]; // If sum drops below zero, reset sum and update start to the next station if (sum < 0) { sum = 0; start = j + 1; } } return start % n ; //helps to avoid error if the starting point is the last index . } }

ayushi didi me abhi 3rd semester me hu.mene abhi dsa padhna start kiya h me bhi aapke jaisa ek achi company me job karunga.....thank you didi for your videos me aapko har jagah follow krta hu.aaap meree liye motivation h bdi behen jaisi

Super

Lld

Mam I want referral.... please can you help me if possible 🙏🙏

chumma pdati ho app.

Hlo didi, I am a 2nd year diploma student in mechanical engineering, main bhi vlogging karna chahta hu kyuki, mera confidence kuch acchha nahi hai so main improve karna chahta hu khudko, agar aap aur koi bhi jo ye comment padhara hai to kuch advice dena yr plzzz, waise mere paas kuch instruments bhi hai hai lekin fir bhi karna chahta hu.

Hi ma'am, your knowledge is amazing. Can you make a video for college students like us for internhsip? I saw Deconstruct 1Lakh internship. Please explain more

Link

Lld

I have taken your 1:1 session I paid the amount..you didn't even join the meet I'm still waiting This is not fair I kept a lot of hopes Needed your guidance Even after payment you are not showing up I don't even have words

Hi mam walmart me kisiko milne ke liye aa sakte he ya appointment lena padta he mam? Matlab unown person ko andar kisiko milneko aloud he mam?

Hi mam walmart me kisiko milne ke liye aa sakte he ya appointment lena padta he mam? Matlab unown person ko andar kisiko milneko aloud he mam?

Best video dosttt 🎉

I always find these backtracking recursion based problems hard to visualize, but your detailed black board explanation and walk-through makes it much easier for me.

Yeah, these combinatorial problems are little challenging to understand, especially with recursion involved. But you have done a great job walking us through the steps patiently before getting into actual code. Nice. 😄

The reason why we are using the min heap because we want to pop the less frequent number not because its nlogn. Also if the question is asking for top less frequent number then we would use max heap because when we pop max heap the max frequent number will be popped