Lecture21: Solving LeetCode/CodeStudio Questions [Arrays]

[@CodeHelp]

Do Visit Relevel: relvl.co/2smk

[@ShouryaPant]

Bhaiyaa notes ki link galt dal gyi

[@Góne-y2s]

reach++

[@CodeHelp]

@@ShouryaPant corrected

[@ShouryaPant]

@@CodeHelp thanx Bhaiyya You are Doing An Amazing Job. 👏 This Course Is Fantastic And awesome. KEEP it Up The Good Work🙂🙂

[@siddharth7261]

@@CodeHelp Sir, I can create an interactive blog that will help students to get topic wise notes along with code for this DSA course.

[@shreyanshthakur5405]

"Programming pe kam focus karo, engineering pe zyada focus karo" (Before bursting out to code, first break the problem into subparts or conditons). Awesome lecture bhaiya

[@deeppatel0586]

if you hate the subject, its because of the teacher. if you love the subject, its because of the teacher. tremendous thanks babbar sir.

[@mdshahidansari9126]

Absolutely

[@saignaneswarsutrave1171]

My approach for Rotate Array problem: 1. Reverse the whole vector 2. Reverse the first k elements of the vector 3. Reverse the remaining elements of the vector code: void rotate(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()); } Dry run: arr = {1, 2, 3, 4} k = 2 step 1: {4, 3, 2, 1} step 2: Reverse first two elements -> {3, 4, 2, 1} (k = 2 -> reverse elements present at 0th and 1st index) step 3: Reverse the remaining elements -> {3, 4, 1 , 2}

[@ashish7604]

hey did you get somewhere or this was your organic thinking if it was what was your approach

[@_PRANAYMATE]

@@ashish7604 Nice bro

[@yatharthahuja1635]

Question 3) Add 2 arrays can also be done in the following way. vector reverse(vectorv){ int s = 0; int e = v.size()-1; while(s=0;i--){ res1 = res1 + (a[i]*mul); mul=mul*10;// this was the changing step } mul=1; for(int i=m-1;i>=0;i--){ res2 = res2 + (b[i]*mul); mul=mul*10; } int total = res1+res2; while(total > 0){ int ele = total % 10; ans.push_back(ele); total = total / 10; } return reverse(ans); }

[@er.preetiyadav1688]

"why I started...?" This line motivates me a lot... Thnks bhaiya..❣️❣️

[@PRIYANSHUCHAKRABORTY-yv3mq]

yes,me 2

[@sounaksaha1455]

Present Bhaiyaji, Aap sirf video dalte rahiye .. humare taraf se full support aur mehenat hum darshaate rahenge... Aur ye course duniya ka best course hai.

[@zhrrBro]

Homework | Time Complexities | 1> TC = O(n) SC = O(n) 2> TC = O(n) SC = O(1) 3> TC = O(n+m) SC = O (n+m) Awesome Lecture !!

[@amitprakhar4802]

Why it is showing time limit exceeded the find sum question

[@amitprakhar4802]

In gfg ide

[@OmPrakash-vk4ko]

ig both time and space should be ----> O(min(M,N) + {max(M,N)-min(M,N)} + 1) becz 1st loop will run till one of the array ends 2nd will run till other part and carry could not be of more than one digit Am I thinking right🤔

[@bcs_Jaskaran]

Check sorted and rotated : { int n = given.size(); vector temp(n); temp = given; sort(given.begin(), given.end()); vector check(n); for (int k = 0; k < n; k++) { for (int i = 0; i < n; i++) { int pos = (i + k) % n; check[i] = given[pos]; } if (check == temp) { return 1; } } return 0; }

[@avijeetpaul1378]

love your channel sir, hats of to you. i have a small add-on to the last question, as i tried to solve the last question before hand of watching the solution, so i came up with my solution as: I would first convert the arrays into an integer, and the sum the both converted integer to get the final answer and then mod the answer with 10 to get the remainders and push that remainder into a vector and later reverse it to get the final answer. Code: #include vector findArraySum(vector&a, int n, vector&b, int m) { int sum1=0, sum2=0; int sig=0; for(int i=n-1;i>=0;i--){ sum1=sum1+(a[i]*pow(10,sig)); sig++; } sig=0; for(int i=m-1;i>=0;i--){ sum2=sum2+(b[i]*pow(10,sig)); sig++; } int sum=sum1+sum2; vector arr; while(sum!=0){ int temp=sum%10; arr.push_back(temp); sum/=10; } reverse(arr.begin(),arr.end()); return arr; } And thank you so much for your incredible persistence sir.

[@vishaljoshi-v5i]

My Approach for the first question is --> void rotate(vector arr, int key) vector::iterator it = arr.begin() + key - 1; reverse(arr.begin(), arr.end()); reverse(it+1, arr.end()); it = arr.begin() + key; reverse(arr.begin(),it);

[@nischalgupta8050]

I found your playlist on dsa few days ago and I loved this playlist

[@sumitshetty295]

Another solution for Array rotation : (Using Reversal Algorithm) void rotate(vector& nums, int k) { int pos = nums.size()-(k%nums.size()); //reversing last k elements reverse(nums.begin()+pos,nums.end()); //reversing remaining nums.size()-k elements reverse(nums.begin(),nums.begin()+pos); //reversing whole vector reverse(nums.begin(),nums.end()); } Time Complexity : O(N) Space Complexity: O(1)

[@LootCrack]

"check if array is sorted and rotated" Sir is question me loop i=1 se i

[@akashthoriya]

me Saturday-sunday pura din beth ke, video dekhta hu aur practice bhi karta hu, Thank you bhai💛🤗

[@cr7motive720]

Blessings of many students are with you Keep going bhaiya 🙏🙏(respect)

[@sunnykakrani7830]

Simple Approach in array sum : we can avoid the usage of carry as well . vector findArraySum(vector&a, int n, vector&b, int m) { int num1=0; int num2=0; int j=n-1; // create the first number for(int i=0;i

[@discoverComputerprogramming7]

bool check(vector &arr){ int count =0; for(int i=1; iarr[i] % arr.size()) count++; } return count

[@pratyush2331raj]

Time Complexities Q1- TC = O(n) SC = O(n) OR // For recursive approach TC = O(n) SC = O(1) Q2- TC = O(n) SC = O(1) Q3- TC = O(max(n, m)) SC = O(max(n, m))

[@shifteditz06]

Question 3 : easiest and shortest solution string s1, s2; for(int i=0;i integer to string vector arr; for(int i=0;i

[@amitshukla2268]

My approach for 3rd question :- vector findArraySum(vector&a, int n, vector&b, int m) { int x=a[0],y=b[0]; for(int i=1;i

[@anuragraut5101]

this was the first approach that came to my mind after seeing the question

[@bishalchatterjee745]

I did a similar code but was facing issue with the carry part.. Thanks for the help buddy>>!!

[@amitshukla2268]

@@bishalchatterjee745 welcome bro :)

[@taranjotsingh2374]

Thanks for the solution. @Amit Shukla but what would be the time complexity of this solution?

[@bishalchatterjee745]

@@taranjotsingh2374 O(n)

[@aniket_k720]

Love bhaiyaa 🌹❤️ Watching your videos since i was in class 11th .....now i am in college and feel.glad that am following ur videos since then.....helpsss too much........🌹

[@abhaysingh1665]

This line "Hello jii this is love babbar" gives us a peace and motivation that i can do 🙏🙏🙏

[@youtubeshortz20]

rotated arrray without using extra space :- k =k%nums.size(); reverse(nums.begin(),nums.end()); reverse(nums.begin(),nums.begin()+k); reverse(nums.begin()+k,nums.end());

[@goldiison9663]

First By the review of your course in KZbin i don't visit your channel but now from this video i recommend my friend your channel.

[@hyderali1520]

bhaiya apka solutions ka approach ko 100 topoo ki salamii.maja araha hai dsa karna ma ab pahla stress ata tha rona ata tha.

[@mahimanbansal1709]

heres another approach for sum of array vector c; int s1=0; int s2=0; for(int i=n-1;i>=0;i--){ s1=a[i]*pow(10,i)+s1; } for(int i=0;i

[@rishabhtyagi7369]

Another solution for finding if the array is rotated or sorted or both: #include using namespace std; void check(int arr[], int n){ int count = 0; for(int i = 0; iarr[i+1]){ count++; } } if(count==1 && arr[0]>arr[n-1]){ cout

[@usersb685]

bhaiya apne ye kaisa logic soch liya mujhe ye dekh ke differentiation yaad aaa gya

[@rahulstudies1178]

q1) time complexity- O(n) q2) O(n); q3) O(n+m);

[@kunalkamthe8039]

This placement series is awesome bhaiya and best ever series on dsa ever...

[@divyanshrathore8878]

My approach for Q2. int count=0; for(int i=0;i

[@varun-qm3xj]

bhaiya aj 3no question khud se lagaye maza he aagya. confidence next level hai ab

[@siftainejaz4772]

Bhaiya.. Sum of two arrays ka aur tareeka hai. I thought it might be helpful for a lot of peaple. So, I am posting this. vector c; int sum=0; int sum1=0; for(int i=0;i

[@devanshraghuwanshi710]

bhai while loop kese kaam kar rha hai yaha pe aap har total ki last digit ko vector ek beginning me dalte jaa rhe ho isse to vector me total ka reverse aaa jayega

[@jayant-baid]

Another Approach for Sum of Two Arrays vector findArraySum(vector&a, int n, vector&b, int m) { int sum1=0,sum2=0; for(int i=0;i

[@avibirla9863]

Nice approach 👍

[@jayant-baid]

@@avibirla9863Thankyou

[@xeroday342]

bhai ye hi mene socha tha ,kya ye cases waki bakchodi karni😅

[@patelrajkumarnareshkumar8156]

easy way to solve sum of two array wala problem vector findArraySum(vector&a, int n, vector&b, int m) { // Write your code here. int A = 0 , B = 0 ; for(int i = 0 ; i < n ; i++){ A = A*10 + a[i]; } for(int i = 0 ; i < m ; i++){ B = B*10 + b[i]; } A= A + B; vector sum; while(A > 0){ int mod = A % 10; sum.push_back(mod); A = A / 10 ; } reverse(sum.begin() , sum.end()); return sum; }

[@abhirajvarshney7394]

(Diff approach) Sum of two arrays : vector findArraySum(vector&a, int n, vector&b, int m) { int ans1 = 0, ans2 =0,ans3=0,ans,i; vector c; i=0; while(i

[@hritavsinghsolanki8893]

🔥

[@hritavsinghsolanki8893]

beautiful approach , easy to implement thanks alot keep coding sir

[@akshayjain1332]

(my approach to q3: using some old tricks taught by luv bhaiya ❤️) vector findArraySum(vector&a, int n, vector&b, int m) { int num1 = 0 ; int num2 = 0 ; int sum = 0; vector ans; for(int i = 0;i

[@taranjotsingh2374]

Can you help in explaining the time complexity of this solution

[@AkshayKumar-gm7mx]

your solution is good enough. but don't repeat the step to calculate num1 and num2. Just define a function to do that for u.

[@jiyasharma9225]

second ques using modulus class Solution { public: bool check(vector& nums) { int n=nums.size(); int count = 0; for(int i=0;inums[(i+1)%n]){ count++; } } return count

[@29-rohitkatare3]

problem 1: TC O(n), SC O(n) problem 2: TC O(n), SC O(1) problem 3: TC O(max(n,m)+(n)), SC O(n)

[@iconliving9076]

Best way of teaching. Quick and quirky. Please keep on making such videos.

[@a_28_mayekarprathameshmang88]

1)Rotate array Time Complexity - O(n) Space Complexity - O(n) 2)check sorted and rotated Time Complexity - O(n) Space Complexity - O(1) 3)sum of two arrays Time Complexity - O[max(m,n)] Space Complexity - O[max(m,n)]

[@karanmehra3305]

in question 3 we can use stack for the (sum of two arrays) resultant array . we don't need to write the reverse function . because the stack works on LIFO principal. thank you love bhai

[@princejacker321]

Time Complexities Q1- Time=O(n) Space=O(n) Q2- Time=O(n) Space=O(1) Q3- Time=O(n+m) Space=O(n+m)

[@backbencher304]

n+n kaise hoga bro ak bar explain kr de.....m confuse ho rha hu is question m

[@amandeshpande8981]

This is way better than uploading a 4 hr video In a day and taking break for weeks

[@AkhilBharti-k5t]

For third question you can use this approach too : vector findArraySum(vector&a, int n, vector&b, int m) { vector ans; int temp1 = 0, temp2 = 0; for(int i = 0; i < n; i++){ temp1 = (temp1 * 10) + a[i]; } for(int j = 0; j < m; j++){ temp2 = (temp2 * 10) + b[j]; } int sum = temp1 + temp2; while(sum != 0){ ans.insert(ans.begin(), sum % 10); sum /= 10; } return ans; }

[@sadaf_r]

Question bahot sahi le rahe ho bhaiya...ekdum kadak level k🔥maja aagya

[@I_Anupam_Pandey]

attendance lecture 21 thank you bhaiya for this amazing video aur notes ma lagta ha apke raw video file upload ho gaye haan

[@anmol3]

Runtime 16 ms < 175 ms Solution 3 //Sum of two arrays vector findArraySum(vector&arr1, int size1, vector&arr2, int size2) { int size1=arr1.size(), size2=arr2.size(); vector ans(max(size1, size2)+1,0); int size3=ans.size(); int i=size1-1, j=size2-1, k=size3-1; int carry=0; while(i>=0 && j>=0){ ans[k]=(arr1[i]+arr2[j]+carry)%10; carry=(arr1[i]+arr2[j]+carry)/10; if (carry==1) ans[k-1]=1; i--, j--, k--; } while(i>=0){ ans[k]+=arr1[i]; i--; k--; } while(j>=0){ ans[k]+=arr2[j]; j--; k--; } if(ans[0]==0){ for(int i=0; i

[@kunalhole6911]

Bhaiyaa Time Complexity bhii Code kii discuss karna video may. After coding discuss the complexity of solution so Practise hoti rahagiii dono kiii code kii bhii aur complexity ki bhii.

[@CodeHelp]

will make it as a practice from next video

[@_IT_RajnishKrYadav]

Add 2 array wala ko aise bhi solve kr skte h n - carry ka jhamela hi nai esme 😅😅 int s1=0, s2=0; //s1 = digit of Array1 & s2 = digit of array2 vector ans; for(int i=0; i

[@arungujjar7328]

Bhaiya, i just saw a comment in telegram channel which u shared... ignore that type of people they themselves are chu****...u are doing a great job... thanks a lot bhai for this type of content for free❣️🙌

[@Mb-nr5nz]

for 3rd question we can also use this , i tried this code by myself #include vector findArraySum(vector&a, int n, vector&b, int m) { // Write your code here. int sum1 = 0; for (int i= 0 ; i < n ; i++ ) { sum1 = sum1 * 10 ; sum1 = sum1 + a[i]; } int sum2 = 0; for (int i= 0 ; i < m ; i++ ) { sum2 = sum2 * 10 ; sum2 = sum2 + b[i]; } int sum = sum1 + sum2 ; vector c ; while (sum != 0) { c.push_back(sum % 10) ; sum = sum / 10; } reverse (c.begin() , c.end()); return c; }

[@ritikchaudhary1232]

Rotate ques : Time complexity -> O(n) space complexity -> O(n) , n is size of nums vector. Sorted and Rotated : Time Complexity-> O(n) space complexity -> O(1) Add array : Time complexity -> O(m+n) Space Complexity -> O(m) or O(n). if I am wrong anywhere please correct me.

[@KushalChandar.]

I think Time Complexity-> O(max(m, n)); //only one of 2 while loops will execute //i,e i >= 0 or j >= 0 Space Complexity-> O(max(m, n)); //carry while loop does can actually be made into a if statement, i think carry can only be 0 or 1, correct me if i am wrong

[@ankushladani496]

Abhi aaya hu dekhne ke pehle hi bol diya done..👍👍

[@Risingstar-k6w]

we can solve 3rd question using one more method by if we create a number by accessing the elements from vector and in initital lecture you taught one formula of ans=ans*10+digit it will create a number then simply add them

[@erenyeager6092]

easier approch for Q1. class Solution { public: void rotate(vector& nums, int k) { int n=nums.size(); k%=n; reverse(nums.begin(),nums.end()); reverse(nums.begin(),nums.begin()+k); reverse(nums.begin()+k,nums.end()); } };

[@jagmohanrai2714]

Explanation of sum of array is awesome bhaiya best explanation I did not see anywhere,explanation like yours thanx

[@AdityaKumar-dz1pf]

Before looking to solution for 3rd ques I tried this one and successfully run. vector findArraySum(vector&a, int n, vector&b, int m) { // Write your code here. int sum_a=0,sum_b=0; vector ans; for(int i=0;i

[@Avi058]

my approach of 2nd ques derived from rotated array ques class Solution { public: bool check(vector& nums) { int n = nums.size(); int i = 0 , j = 0 , k=0; vectortemp(n); for(int i = 0 ; i < n-1 ; i++) { if(nums[i] > nums[i+1]) { k = i+1; break; } } while(i < n) { j = (i + (n-k) )%n; temp[j] = nums[i]; i++; } sort(nums.begin(),nums.end()); if(nums == temp) {return true;} else {return false; } } };

[@rishabhrajdhiman3005]

I did the last one like this : vector findArraySum(vector&a, int n, vector&b, int m) { // Write your code here. int numa = 0; int numb = 0; for(int i=0 ; i

[@shobhitkatiyar9224]

Rotate ques : Time complexity -> O(n) space complexity -> O(n) Sorted and Rotated : Time Complexity-> O(n) space complexity -> O(1) sum of two array : Time complexity -> O(n+m) Space Complexity -> O(n).

[@piyushborkar6803]

Love bhaiya content bohot tagde level ka aa raha hai 🔥🔥😍 Aise hi banate raho

[@valendradangi1822]

We can also do the 3rd question as follows: vector findArraySum(vector &a, int n, vector &b, int m) { int num1 = 0, num2 = 0, num3r = 0, num3; for (int i = 0; i < n; i++) { num1 = num1 * 10 + a[i]; } for (int i = 0; i < m; i++) { num2 = num2 * 10 + b[i]; } num3 = num1 + num2; cout

[@nainaryan]

we have to use ans.pop_back(sum); in 3rd question to avoid reverse function

[@vigneshv7632]

I sticked until last minute. I am liking it. Let's keep the josh high🔥

[@crazy_stanle]

Semester chalrahe hai Semester khatam hote hi Sare vedios dekhlungaa APP DAREHAATE RAHOO BABBAR BRO

[@Shivamkumar-xu6hu]

#consistency op Both your and mine let's see who will break first 😂

[@dbgaming7722]

another approach to solve rotate an array class Solution { public: void reverse(vector& arr, int s, int e){ while(s

[@mitesh3579]

*My approach for "check if array is sorted and rotated..."* class Solution { public: bool check(vector& nums) { int n = nums.size(); int ans =0; for(int i=0; i nums[(i+1)%n]){ ans++; } } if (ans

[@bellalmahtab8220]

Bhaii yhaa num se pehle ampersand kyu lgaya haii btayee ga Kya

[@blackfox3586]

In sorted and rotated how nums[i-1]>nums[i] if it's sorted

[@ayushanand4938]

//Add 2 arrays(Without extra array/vector) vector findArraySum(vector&a, int n, vector&b, int m) { //setting the large array in a and small in b if(n=0){ int sum=a[j]+b[i]+carry; carry=sum/10; sum=sum%10; a[j]=sum; i--; j--; } while(j>=0){ long sum=a[j]+b[i]+carry; carry=sum/10; sum=sum%10; a[j]=sum; if(carry==0) break; j--; } if(carry!=0) a.insert(a.begin(),carry); return a; }

[@OmPrakash-vk4ko]

for third question this could also be one of the solution : - class Solution{ public: vector findSum(vector &a, vector &b) { // T.C & space complexity ----> O(M+N) // finding integers vector ans; int num1=0, num2=0; // 1st number int j=0; for(int i=a.size()-1; i>=0; i--){ int last_D1 = a[i]; num1 += pow(10,j++)*last_D1; } // 2nd number int k=0; for(int i=b.size()-1;i>=0;i--){ int last_D2 = b[i]; num2 += pow(10,k++)*last_D2; } // new num = sum int sum = num1 + num2; while(sum!=0){ int lastD = sum%10; ans.insert(ans.begin(),lastD); sum /=10; } return ans; } };

[@awais_ansarii]

Thank you for this placement series Bhaiya We're learning and enjoying a lot.

[@sujalsharma5181]

its good that you were able to solve this rotated and sorted question but i don't think anyone should approach a question like you did which was depending on the test it worked this time but won't work every time and plus it was very clear that you have already been to this question many times, like the way you approached tells that there was literally no concept used

[@sujalsharma5181]

go and watch NeetCode's solution, the way that guy solved the question is exactly how you should approach and analyze a question

[@sujalsharma5181]

And want to know why he was able to solve this ? because he is very good at this but remember not everyone can teach others

[@patelrajkumarnareshkumar8156]

we can solve rotate array without using extra space class Solution { public: 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()); } };

[@kaushikmandal8061]

1st question -> TC = O(n), SC = O(n) 2nd question -> TC = O(n), SC = O(1) 3rd question -> TC = O(n+m), SC = O(temp)

[@geekyprogrammer4831]

I like your course more than MIT Data Structures and Algorithms!

[@aditya_jainn]

1) time complexity- O(n) 2) O(n); 3) O(n+m);

[@MOTO-SP]

my apporach for q3 eg first we take a[ ] = { 4,5,1} and b[ ] = { 3,4,5} we add a as 451 by (ans x 10) + digit ( i.e here 4) == 4 now ( 4 x 10 ) + 5= 45 (45 x 10) +1 = 451 similarly we get 345 now we add them to get 796 now Get the last digit of the number Insert the digit at the beginning of the vector number /= 10; // Remove the last digit from the number to get c[ ] ={7,9,6} open for suggestions

[@apurbakumarmajumder9478]

Love from Delhi bhaiya. Best DSA series.

[@bansalshivam]

Reverse array algo to rotate the array is better than modulo algo as in-place solution is require and it has better space complexity O(1).

[@prasadprashantb.4001]

Bhaiyaa last ka code thoda nhi smja.. Parr phle do achhe the 🙌🤘🤘❤❤thanks bhiyaa👏👏🙌🤝nice session on leetcode..

[@sanskargour6673]

22:16 Question 3 (alternate code) - #include using namespace std; int digit (int arr[],int n) { int digit = 0; for (int i = 0 ; i < n ; i++ ) { digit = digit* 10 + arr[i]; } return digit; } int main() { int arr1[3] = {1,2,3} , arr2[2] = {9,9}; int n = 3 , m = 2 ; cout

[@prakharlowanshi1781]

good bro nice solution

[@sanskargour6673]

@@prakharlowanshi1781 tq

[@poojayadav695]

Ques1-Time complexity-O(n) space complexity-O(n) Ques2-Time complexity-O(n) space complexity-O(1) Ques3-Time complexity-O(n+m) space complexity-O(max(n,m))

[@priyanshubari9685]

on leetcode, the first question states that you have to solve it using constant space complexity. the solution provided has linear space complexity.

[@9852963775aaa]

bhaiya teeno questions solve hogaye thankyou so much bhaiya>>>>>>

[@sagarnivargi9202]

This playlist is best in the India

[@ritangshudasgupta6516]

Bhaiya ye approach try kr skte h last question k liye? int x=0; int y=0; for(int i=0 ; i

[@coder9760]

ADD ARRAY another Aproach :- vector findArraySum(vector&a, int n, vector&b, int m) { int num1=0,num2=0,ans; vector temp; for(int i=0;i

[@programmer545]

ques 1 : time complexity : o(n) space complexity : o(n) ques 2: time complexity : o(n) space complexity : o(1) ques 3: time complexity : o(n+m) space complexity : o(n+m) please correct if wrong anywhere?

[@sandipdegloore3494]

Time complexities : 1} time = O(n) space complexities = O(n) 2} time complexities = O(n) and space complexities = O(1) 3} time complexities = O(n) or O(m) and space complexities = O(n)

[@BEEMonikaBishnoi]

Abhi nhi dekh pa rahi but jarur dekhungi piche wala cover kr leti hu

[@amitshukla2268]

bhaiya thnx a lot , attendance marked, bss bhaiya eise hi 17 March tak course khtm krrdo . 🤗🔥

[@dwivedi6037]

Rotate arrays: time complexity ==> O(n) space complexity ==> O(n) , n is size of nums vector. Rotated and sorted array: time complexity ==> O(n) space complexity ==> O(1) Add array : Time complexity ==> O(m+n) space complexity ==> O(m) or O(n).

[@gurpreetSingh-ir3ho]

rotate an array----> class Solution { public: 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()); } };

[@shivamchaudhary7844]

my approach for 3rd question is :- vector findArraySum(vector&a, int n, vector&b, int m) { vectorans; int i=n-1,j=m-1,sum=0; while(i>=0&&j>=0){ sum+=a[i]+b[j]; if((sum%10)==sum){ ans.insert(ans.begin(),sum); sum/=10; } else{ ans.insert(ans.begin(),(sum%10)); sum/=10; } i--; j--; } while(i>=0){ ans.insert(ans.begin(),sum+a[i]); sum=0; i--; } while(j>=0){ ans.insert(ans.begin(),sum+b[j]); sum=0; j--; } if(sum!=0){ ans.insert(ans.begin(),sum); } return ans; }

[@AakashShah-y3s]

Ek he to dil hai , kitne bar jitoge babbar bhai

[@DataStructuresDecoded]

much easier way to solve question 3: void sum(int arr[],int brr[]){ int x=0,y=0;int sum=0; //initializing some variables vector a; //vector to store the answer for(int i=0;i

[@RohitRana-tz2lr]

Thanks, bhaiya, I am continuously watching your DSA series. I am loving it so much

[@thoyazpc7442]

my Approach for : Sum of two arrays " " " vector findArraySum(vector&a, int n, vector&b, int m) { // Write your code here. vector ans; int sum1 = 0; for(int i=0 ; i