Rotate Array by K places | Union, Intersection of Sorted Arrays | Move Zeros to End | Arrays Part-2

[@takeUforward]

Let's march ahead, and create an unmatchable DSA course! ❤ Timestamps someone please :) Use the problem links in the description.

[@brajeshmohanty2558]

At 23:47 my mind just got blown away. Me be like: - itna easy kese Benstocks 😂🙏. Bhai iPad ka notes ka bhi pdf dalo na . Tumaaher writing sei sab samajh mai ata hai. Boh tuf sight par itna bada article dekhne sei dar lagta hai 🙂. But you are doing amazing job bhai thank you 🙏😌.

[@Atharva_kathiyawadi]

appreciable !!

[@ParasSharma-mf8or]

Please upload the next video.

[@gautamsethiya1397]

@takeUforward class Solution { public: void rotate(vector& nums, int k) { int n = nums.size(); k =k%n; int temp[k]; for (int i = n - k; i < n; i++) { temp[i - n + k] = nums[i]; } for(int i=n-k-1;i>=0;i--) { nums[i+k]=nums[i]; } for(int i=0;i

[@yaswanthmitta8983]

for moving zeroes to the end we cen write it in 1 loop instead of two though time complexity is same , complexity for understanding decreases : v oid moveZeroes(vector& nums) { int j=0; for(int i=0;i

[@zenithria_414]

Who is grinding their head with DSA in their summer vacation?? Heartly thanks to striver for making us understand dsa in such depth :)

[@prasannagodiyal6736]

Me😃

[@avgguy7129]

me too lol

[@JohnCena-uf8sz]

bro this course from his website is completely free for now right ? ... or it has premium paid version as well ?

[@victory7361]

What do u mean by summer vacation 😢?

[@AdityaKumar-gx2ip]

For Right Rotation of Array void rotate(vector arr, int d) { if(d>size) d = d % size; reverse(arr.begin(),arr.end()); reverse(arr.begin()+d,arr.end()); reverse(arr.begin(),arr.begin()+d); }

[@bishalkundu7592]

Thanks a lot sir for taking out time from your busy schedule and making videos and uploading those videos in the span of 2-3 days. It takes a lot ot effort. Thanks a lot 😊😊

[@arnab_16]

not only your dsa knowledge is good also your communication skill, voice tone , body movement is superb 🥰🥰

[@sanjana6931]

Agreed. He's so smart and confident, definitely a role model to students!

[@luckshaeey]

27:00 rotating an array left by 1 place is equal to rotating it right by n-1 places. We can use this information for rotating an array to right by d places.

[@takeUforward]

Perfect thought!

[@bhargavanand7325]

great

[@himanshukaushik9223]

Bhai leetcode pa left rotation question reverse .end() and begin() use karna pad Raha hai but gfg ma nahi why

[@bhargavanand7325]

@@himanshukaushik9223 in case of vector we use stl function like begin() and end() but in case of array we don't use it for example in gfg

[@himanshukaushik9223]

@@bhargavanand7325 thanks bro

[@ruhiawasthi9]

I have written this approach for rotating the array to the right by k steps : void rotate(vector& nums, int k) { k%=nums.size(); reverse(nums.begin(), nums.end()); reverse(nums.begin(), nums.begin() + k); reverse(nums.begin()+k, nums.end()); }

[@himadriroy7980]

I did not only enjoy the lecture but loved it too. Thanks Striver Bhaiya for explaining so lucidly. ❤

[@hey-cg3fv]

Move zeroes question another approach with O(1) space int j=0; // Firstly taking all non- zeroes number for(int i=0;i O(1)

[@ByteBuilder-b6u]

Really your thought process is really clear and the way you write the code is awesome.For union of two arrays you have written the code in such a simpler way whereas in first go I wrote the code this way : vector < int > sortedArray(vector < int > a, vector < int > b) { // Write your code here int i = 0; int j = 0; vector ans; while(i < a.size() && j < b.size()){ if(a[i] < b[j]){ if(i == 0){ ans.push_back(a[i]); } else if(a[i] != a[i-1]){ ans.push_back(a[i]); } i++; } else if(a[i] == b[j]){ if(i == 0){ ans.push_back(a[i]); } else if(a[i] != a[i-1]){ ans.push_back(a[i]); } i++; j++; } else{ if(j == 0){ ans.push_back(b[j]); } else if(b[j] != b[j-1]){ ans.push_back(b[j]); } j++; } } while(i < a.size()){ if(ans.back() != a[i]){ ans.push_back(a[i]); } i++; } while(j < b.size()){ if(ans.back() != b[j]){ ans.push_back(b[j]); } j++; } return ans; }

[@honblegaurav9917]

can you tell me why i++ and j++ should be outside of if else loop. because for me it is not accepting the solution but accepts when i put i++ and j++ inside the if else loop

[@Str13011]

Thank you for all the tutorials, Please keep supporting from starting to get good placement by the help of your tutorials!

[@puneetnj]

Right Shift Solution(Leetcode 189): k = k % arr.length; reverse(arr, 0, arr.length-1-k); reverse(arr, arr.length-1-k+1, arr.length-1); reverse(arr, 0, arr.length-1);

[@hari7939]

Just do k=k%n, k=n-k!! as simple as that

[@senseiAree]

I always thought that the optimal solution to this code was the hashset one that you did in the middle part. It blew my mind when I saw the optimal one... And that too so easy man... Wow .. just wow to your observation ❤

[@codeby_chance6728]

Striver please bring string series. Much needed

[@sahadevbhaganagare4761]

me first getting the optimal solution and then finding the brute 🤣🤣

[@he_KARTIK]

mee too😂

[@Vikas-dm9uc]

#me2 moment

[@helloworld-n1z]

how to remember logic ,i always forget nd cant able to code

[@sahadevbhaganagare4761]

@@helloworld-n1z I am not the right person but still always go back to the code after a week or so and try solving it again. Then review it after a month. That is it, it will get hard coded in your brain after that.

[@Me_Meraj]

Bro me too 😢 Isi wajah se easy problem solve krne me v time lgta h 😂😂 dimag me bs chlta h ki optimal hi krna h

[@AbhishekSingh-rp6hp]

Your course is awesome wish I could have found your course a year before currently I am in 3rd year of my clg 😊

[@anshu59kumari]

From which college u r

[@AbhishekSingh-rp6hp]

@@anshu59kumari Techno International New Town West Bengal

[@sanjaykrishna689]

left rotation to right rotation: simple approach is to think oppositely to left rotation. Shift all elements right of d and add to temp and push it. Just reverse the order of the reverse statements used in left rotation!.

[@MotivexBeastt]

Striver is the lion of this community

[@MONESHSG]

Brother, Actually only because of you i came to know what is coding and how to perform our own logic while writing code. Your teaching was nice and I really love it. once before viewing your videos i may code but i don't know how it works internally, but after seeing ur videos it is very easy to understand the working ❣. my first own logic: vector temp; int first = arr[0]; for(int i=1; i

[@kushagramishra5638]

Understood 😄. Logic used for right rotate the array k places is that instead of rotating array right by k places I have rotated it n-k places to left . so code for this is - class Solution { public: void rotate(vector& nums, int k) { int n = nums.size(); k = k % n; reverse(nums.begin(),nums.begin()+(n-k)); reverse(nums.begin()+(n-k),nums.end()); reverse(nums.begin(),nums.end()); } };

[@Miracle_shaw]

hey, me also applied same approach

[@itzmartin20]

same here

[@jatinukey4062]

Exactly right

[@expl0itr]

reverse(nums.begin(), nums.end()); reverse(nums.begin(), nums.begin() + k); reverse(nums.begin() + k, nums.end()); u can do this also.

[@tushartyagi4965]

Rotate right ---->>>> void rotate(vector& nums, int k) { vector temp(nums.size()); for(int i=0; i< nums.size(); i++) { temp[(i+k)%nums.size()] = nums[i]; } nums=temp; }

[@AniketKumar-tc8cz]

Timestamps: 0:00:00 - Intro 0:00:55 - Q1. Left rotate an array by 1 place 0:07:33 - Q2. Left rotate an array by D places 0:26:36 - Follow up Right rotate an array by D places 0:27:11 - Q3. Moving Zeroes to end 0:40:54 - Q4. Linear Search 0:43:01 - Q5. Union of Two Sorted Arrays 0:59:04 - Q6. Intersection of Two Sorted Arrays 1:12:22 - Outro

[@MAX_RITIK]

at 39:57 int j=0; for (int i = 0; i < n; i++) { if(arr[i]!=0){ swap(arr[i],arr[j]); j++; } for moving zeros to end also o(n)

[@md.ualiurrahmanrahat2400]

Thanks for providng such amazing course for completely free. We are intended to you.

[@shreyasr2885]

he never skips anything, Patience is next level!!

[@UECSoumyaRay]

code for right-rotation: void rotate(vector& nums, int k) { int n = nums.size(); if (k > n) { k = k % n; } reverse(nums.begin(), nums.begin()+n-k); reverse(nums.begin()+(n-k),nums.begin()+n ); reverse(nums.begin(), nums.begin()+n); }

[@crazybro4383]

Hii brother can you pls explain me why time complexity of intersection is O(1) and not O(n), where n is the size of larger sized array. Time stamp- 1:12:01

[@architha4305]

for the first time i guessed which concept can be used it might sound like a small thing but as a beginner who is just learning to code actually implementing codes without any reference and knowing which loop is used is really HUGE for me. THANK YOU SOO MUCH

[@zenmonk29]

same for me. feels like a big leap

[@KrishnaKumar-b4m9p]

hi in which question u implemented the code yourself? i am having a hard time in writing code for right rotate an arrray ky k places on my own...

[@satishmaurya7032]

What a wonderful video, it was vey good. Your way of teaching concepts through questions is best.

[@devanshimehta780]

my motivation are your efforts, wont let them go in vain, thankyou STRIVER

[@aswinsrikanth408]

Understood and what an explanation Love your explanation bro, just continue doing what you are doing for the coding community Amazing person

[@curs3m4rk]

I recommend everyone to solve the linked leetcode problems attached as well, as the code needs to be tweaked little bit . So u can understand and do by yourself Great approaches Striver. thanks

[@valiant4498]

Guys i just wanna ask something....r u guys doing all these problems on your own or like just watching the lectures

[@ishanmoykhede9484]

First Understanding the problem then finding approach to solve it then if I make the correct solution then codding it in vs code and moving to nxt question and if I didn't get the solution then watching striver

[@AbhinavTiwari-u7b]

done the assignment and understood each and every concept . Sir appreciate your hardwork

[@sikanddhingra5835]

bhai ki pehli video lgayi h ye mene. bohut maja aya, dil garden garden hogya bilkul. thanku bhai.😁

[@vasudev6559]

One more approach can be: void moveZeroes(vector& nums) { int n=0; for(int i=0;iO(N) SC->O(1) similar but less code

[@kirankawade3424]

Thank you so much for sharing the content free of cost!! very nicely explained too😘

[@RahulKumar-bd6py]

Your are best amongst all I found on internet hatsoff to you sir.

[@yashsaxena7787]

Great! btw in the worst case of Intersection, the space complexity would be O(n1>n2?n2:n1) because whichever one is smaller matches completely with the other array and ends. P.S.- the space is for returning the answer and not solution.

[@b.bhargavitejaswi7655]

best teacher i have ever found in internet

[@uttamvashisth4072]

Salute to this consistency.

[@rohandhalpe9084]

😆😆

[@saurabhtiwari9614]

@@rohandhalpe9084 😡😡😡

[@rohandhalpe9084]

@@saurabhtiwari9614 kya hua bhai 😔

[@saurabhtiwari9614]

@@rohandhalpe9084 Bhai aaisa face kyo bnaya h.

[@PonVignesh-gv1xp]

Learning a lot outta this guy..

[@chiragbirla5606]

One thing striver..... In the sheet for the same question as we know there is a link attached for the gfg version and for the leetcode version but in some cases leetcode version is harder and sometimes gfg version. So please pick the hard problem for that

[@ammarshaikh4054]

if you are geting confused in last loop at 18:55 i.e for(int i=n-d; i

[@siddiqabr7110]

Bro the answer is fkn wrong has no one noticed it at all

[@ammarshaikh4054]

​@@siddiqabr7110Yes bro, i also got it wrong, when referred from video!

[@AdityaKumar-be7hx]

Actually, there is another optimal solution for moving zeroes problem. It is similar to the function we use in quicksort algorithm (quickselect part for arranging elements around pivot). void Optimal2(̀vector& nums) { int j=-1, n=nums.size(); for(int i=0; i

[@nagame859]

Exceptionally concise👏

[@KrishnaKumar-b4m9p]

can u explain how the swap function works here. we want to move all the zeroes in the end how nums[++j]; ????

[@KrishnaKumar-b4m9p]

u wanna see my solution?? void Zero_at_last(int arr[], int n){ int i=0; for (int j = 1; j < n; j++) { if(arr[i]==0 ){ if(arr[j]==0){ continue; } swap(arr[i], arr[j]); } i++; // cout

[@dhruvpatel0503]

In move zeroes to end problem we can use this also..., int y=0; for(int i=0;i

[@anilkumar4200]

Why bro Why, Brute force->better-> optimal, Its really a big effort. thanks for such wonderful video series. Love the way u explain.

[@triptisharma5550]

Your videos are awesome. I am finally learning to think about a problem and how to approach it. Thank you so much ❤️😬

[@thebishalpaul]

For left rotate by D places Brute can be : TC: O(n^2) if d==n Hence the rest two solutions shown in video will be better and optimal. int n=arr.size(); while (k>0) { Integer temp=arr.get(0); for (int i = 1; i < n; i++) { arr.set(i-1, arr.get(i)); } arr.set(n-1, temp); k--; } return arr;

[@sangeetasharma8574]

Thank You Striver for such an effort. Your explanations and teaching is really amazing. Its really Valuable /priceless content :)

[@AnshulVerma-cd1od]

Thank you for making such great and useful videos by taking out time from your busy schedule.

[@anuragprasad6116]

Concise solution for move zeroes to end: int p = 0; for (int i = 0; i < n; i++) if (arr[i]) swap(arr[p++], arr[i]);

[@LearnwithEase20]

Love your explanation sir!You always make things simpler

[@chandusweety3341]

class Solution { public void rotate(int[] nums, int k) { int n=nums.length; k=k%n; reverse(nums,n-k,n-1); reverse(nums,0,n-k-1); reverse(nums,0,n-1); } void reverse(int arr[],int start,int end) { while(start

[@tanishagarwal5992]

In the three reversals, we are doing iterations just for half the time i.e O(d/2) + O((n-d)/2) + O (n/2) making it O(n) so TC as well as SC of optimal solution is better than the better solution.

[@chidambarjoshi3470]

In the brute force approach of union of arrays, you could have directly returned the set instead of taking a temp vector and copying it all over, because the set would only take the single copies of the elements are the time of insertion

[@nashimbiswakarma]

bro you are right but most of the time they i.e., dsa practicing websites expect you to return the result in the form of array(for c++ vector and for Java ArrayList).

[@prateekbhaisora]

void rotateright(vector& nums, int k) { k = k % nums.size(); reverse(nums.begin(), nums.end()); reverse(nums.begin(), nums.begin()+k); reverse(nums.begin()+k, nums.end()); }

[@harshavardhan184]

This much consistency is all we needed bhaiya🤞🤞🤞kudos to you

[@karthikjadhav346]

I guess I have an approach for "better" solution for "Move zeroes to end" ( 27:13 ) , correct me if i'm wrong but this is what I got, with T.C O(N+C) int count = 0; for(int i=0;i

[@rohandhalpe9084]

Bhaiyaa ji please upload videos as fast as possible .Our placement season starting soon in June.💫💫💫💫💫💫💫💫💫💫💫💫💫💫

[@Hi_kartik]

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative. class Solution: def rotate(self, arr: List[int], d: int) -> None: n = len(arr) d%=n def reverse1(i,j,arr): while i

[@tarun1684]

I gonna fainted..after watching at last that this was easy section..what will happen god in future..bcz I m beginner and finding difficulties 😢😢

[@nishantsindhu3420]

Me too bro similar situation 😢

[@sirishayatham1810]

You are making easy for us to understand DSA!! Heartfully Thanks to Striver Brooo

[@ParasSharma-mf8or]

0:00 Introduction of course 01:01 Left rotate an Array by 1 Place 04:38 Pseudo code 07:17 Code-compiler 08:05 Left rotate an Array by D Places 14:38 Pseudo code 20:18 Code-compiler 27:13 Move zeroes to end 37:41 Pseudo code 40:30 Code-compiler 41:05 Linear search 42:01 Pseudo code 42:43 Code-compiler 43:04 Union of two sorted array 53:33 Code-compiler 59:05 Intersection of two sorted array 1:09:30 Code-compiler

[@takeUforward]

Thank you Paras!

[@ParasSharma-mf8or]

​@@takeUforward Welcome and thanks for this amazing course.

[@samarthpai5359]

for rotaing by right: void rotateR(vector&arr, int k){ int n = arr.size(); reverse(arr.begin(),arr.begin()+n-k); reverse(arr.begin()+n-k,arr.end()); reverse(arr.begin(),arr.end()); }

[@Lucifer-xt7un]

Bro requesting you please upload videos alternatively as promised bcoz completely relaying on you for placements🥺🥺

[@NithinvKumar-uk2he]

Yes you are right, striver bro please we are requesting you please upload the video as promised by you 🙏

[@takeUforward]

I am keeping my promise, if you see am covering 5 problems in one video. So recording such a long video, then editing it, then uploading it takes time. I can upload daily, if you want me to to cover one video.

[@NithinvKumar-uk2he]

@@takeUforward ok brother thanks for helping us .we are waiting for upcoming videos ☺️

[@harshavardhan184]

True

[@Lucifer-xt7un]

@@takeUforward understood ur pain bro thanks for everything.

[@Gokul-gkl]

We will have 2 solution one is talking the first 3(in case of d is 3) elements in the temporary array and then making the last 4 elements to the left most and then put back the first 3 elements in the temporary array to the array as you did And the another solution will be FIRST storing the LAST FOUR(which means for(i=d; i

[@subhajitmandal1869]

"THERE IS A ONLY ONE KING IN DSA"

[@Mythri333]

Really it is world biggest best dsa course. 😊😊

[@SatwikSharma21bcs080]

✦ Learn left rotation of an array by one place 00:03 ✦ Understand the difference between space used and extra space in algorithms. 06:07 ✦ Learn how to left rotate an array in C++ or Java 18:42 ✦ Left rotate an array by D places 24:15 ✦ Use two pointer approach to move zeros to the end of an array. 34:25 ✦ Summary of array problems 40:19 ✦ Implementing a pointer approach to find the union of two sorted arrays. 50:18 ✦ Union and intersection of sorted arrays 55:30 ✦ Optimize time complexity using two pointer approach 1:06:02 ✦ Optimal solution for intersection of two sorted arrays Click to expand 1:11:14

[@Unstoppable.Sakshi]

I have reached here , sir . By following your A to Z DSA course . Thanks a lot sir..

[@bipinsingh1490]

Can u tel me is that a fee course or free course?

[@Akash-yr2if]

Hey Striver, Won't this work fine for the Move Zero's to the end : void moveZeroes(vector& nums) { int i=0; for(int j=0;j

[@arunimmalviya1605]

How do you know if the first index i.e. i=0, will always going to be zero? Also you cannot initiate j from 0, if arr[] = {1,2,3,0}, both I and j will point to 1 and since they both are non zero elements, you are swapping them with themselves only.

[@dhruvsolanki4473]

Clear explanation and amazing teaching.

[@vivek5144]

Hi Striver can you please please upload the notes which you are explaining it will be very helpful 🙏🙏

[@RamPageMMA]

Take your own notes

[@eakanshbhardwaj]

43:50 its not i

[@suriyavs8685]

Hey striver wonderful video. I have a query. The code will be applicable only if the vector or array is sorted in the intersection part. So do we need to sort it again or is there any other optimal approach ?

[@VaishaliDhokchawle]

Thanks a lot sir for taking out time from your busy schedule and making videos. It takes a lot of effort. Thanks a lot.

[@sudhirdaga3453]

Right rotate array by d places is equivalent to left rotate array by n-d places. void rightRotateArrayOptimal(int arr[],int n,int d){ d = d%n; d = n-d; reverse(arr,arr+d); reverse(arr+d,arr+n); reverse(arr,arr+n); }

[@ayushmittal9666]

It's wrong it should be d=n%d reverse(a,a+n) reverse (a,a+n-d) reverse (a+n-d,a+n)

[@rainbowwords6931]

void rotate(vector& nums, int k) { int s = nums.size(); k = k % s; if(k == 0) return; reverse(nums.end()-k,nums.end()); reverse(nums.begin(),nums.end()-k); reverse(nums.begin(),nums.end()); }

[@itz_fuzzy6941]

Munnawar Faaroqi explaining dsa questions...😂😂😂😂😂

[@hidden_star14]

??

[@DeepakKumar-e2s5q]

Aisa hai kaide me raho

[@itz_fuzzy6941]

@@DeepakKumar-e2s5q Ooye Chomu😂 tujhe pta to hai kya bolna chaha rha h tu

[@AnasKhan-pm5ys]

vector rotateArray(vectorarr, int k) { reverse(arr.begin(),arr.begin()+k); reverse(arr.begin()+k,arr.end()); reverse(arr.begin(),arr.end()); return arr; } Striver check it !

[@aryankoul1238]

Not good way to in interviews rounds

[@rish7394]

Bhaiya kab tk poora hoga 🥲

[@bishalkundu7592]

In a previous post bhaiya said that he will try to cover this course in the next 3 months.

[@nikhild.20]

Reverse Approach Code for Right Rotate:- class Solution { public: void rotate(vector& nums, int k) { int n = nums.size(); k=k%n; reverse(nums.begin(),nums.begin()+n-k); reverse(nums.begin()+n-k,nums.end()); reverse(nums.begin(),nums.end()); } };

[@crazyxyzk143]

// right rotation by k places int k;// no of rotation cin>>k; int d = k%n; reverse(arr,arr+(n-d)); reverse(arr+(n-d),arr+n); reverse(arr,arr+n); for(int i: arr){ cout

[@PrithaMajumder]

Raj, Thanks a lot for This Amazing Video about C++ Arrays Video - 2 Completed ✅

[@udaytewary3809]

we can slightly optimise shift 0 to end by using only 2 loops that by first we will construct the temp array as bhaiya created but what additionally we can do is we will maintain a variable count which will store the count of n9.of zeros in the array And then after constructing of temp array we would insert those zeroes at the end of array using a loop or insert as arr.insert(arr.end(), count_zeroes,0) this will insert zeroes at the end of array

[@KiritiSrisai]

your explanation is awesome and it is easy to understand

[@RaunitJaiswal-s9v]

agar kise array ko left rotate kar re hai toh starting from zero index till d index ....and uske baad ka elements ko left lejao and temp array ke elements ko daal do original me ......aur agar kise array ko right rotate kar re hai toh starting from last index till last index-d-1 .....and uske baad ke elements ko right lejao and temp array ke elements ko daal do original me ..........bsc op...raj bhaiya like kardete toh thoda acha lagta

[@lost_hand]

Such a great lecture i wouldn't never forget these conceptss❤❤

[@chinmaymathakari9704]

Thank you so much for taking out time of your busy schedule. Can you please make such videos on Trees as well ? That will be a great help. Thanks a lot😊😊

[@senpaidawwwggg]

two pointer approach is great ...

[@suprithsk9595]

left rotate without extra space: i=0 for j in range(1,n): arr[i],arr[j]=arr[j],arr[i] i+=1 return arr

[@harshitbajeta206]

Thank you for all the tutorials till now and much more waiting for the videos of those questions, which are in the A2Z DSA course sheet.

[@pranshugoyal1571]

Alternative solution to move zeroes to end - void moveZeroes(vector& nums) { int i = 0, j = 0; while(j < nums.size()){ if(nums[j] == 0){ j++; } else{ swap(i, j, nums); i++; j++; } } }

[@ayansiddiqui7383]

can you explain the logic behind this please

[@abhishekjadhav2250]

conscise solution for move zeroes to end class Solution { public: void moveZeroes(vector& nums) { int n = nums.size(); int j = 0; for (int i = 0; i < n; i++) { if (nums[i] != 0) { swap(nums[i], nums[j]); j++; } } } };

[@Codro09]

26:06 for right shift int n = nums.size(); k = k%n; //temp varible to store the values which are n-d from left vector temp; // int temp[]={}; for(int i = n-k;i=0;j--){ nums[j+k] = nums[j]; } //Put back the elements of" temp" int z = 0; for(int l = 0;l

[@sunnymishra2120]

I'm really lucky to have you as a teacher

[@pritishpattnaik4674]

Incase somebody need the code for right rotate , it is in Leetcode rotate array Solution : void rotate(vector& nums, int k) { int n = nums.size(); k = k % n; vector temp; for (int i=n-k ; i=0 ; i--){ nums[i+k] = nums[i]; } int j = 0; for (int i=0 ; i

[@saumyasarngal6109]

Thanku

[@Himani-t3g]

understood! really like the relatable example of relationship🤣🤣 thank you for your efforts

[@vivekbharadwaj4170]

I wish I could get to know about your course in my first year, I'm in second year now, not late, btw thanks a lot (UNDERSTOOD)

[@cinime]

Understood! Fantastic explanation as always, I've enjoyed the video. Thank you very much!!