"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
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}
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); }
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.
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🤔
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
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.
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
(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
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.
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
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());
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);
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........🌹
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++; } }
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; }
//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; }
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❣️🙌
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
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; }
*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
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)]
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).
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.
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()); } };
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;
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
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
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).
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; } };
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
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))
Check if rotated and sorted array - Without extra space ( temp ) --- > O(n) | O(1) solution: class Solution { public: void rotate(vector& nums, int k) { k = k%nums.size(); int s1 = 0, e1 = nums.size()-k-1; int s2 = nums.size()-k, e2 = nums.size()-1; while(s1
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)
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
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
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
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());
@@shubham5934 bhia web development itna important nahi jitna dsa he. Aur waise bhi web d samajh sakte ho apne ap se par dsa samajhne ke liye teacher. So let bhaiya teach dsa first
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
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
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
![Lecture 20: Solving LeetCode/CodeStudio Questions [Arrays]](http://i.ytimg.com/vi/MPvr-LmaZmA/mqdefault.jpg)
Do Visit Relevel: relvl.co/2smk
Bhaiyaa notes ki link galt dal gyi
reach++
@@ShouryaPant corrected
@@CodeHelp thanx Bhaiyya You are Doing An Amazing Job. 👏 This Course Is Fantastic And awesome. KEEP it Up The Good Work🙂🙂
@@CodeHelp Sir, I can create an interactive blog that will help students to get topic wise notes along with code for this DSA course.
"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
if you hate the subject, its because of the teacher.
if you love the subject, its because of the teacher.
tremendous thanks babbar sir.
Absolutely
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}
hey did you get somewhere or this was your organic thinking if it was what was your approach
@@ashish7604 Nice bro
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);
}
I found your playlist on dsa few days ago and I loved this playlist
"why I started...?"
This line motivates me a lot...
Thnks bhaiya..❣️❣️
yes,me 2
Blessings of many students are with you
Keep going bhaiya 🙏🙏(respect)
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.
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 !!
Why it is showing time limit exceeded the find sum question
In gfg ide
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🤔
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
this was the first approach that came to my mind after seeing the question
I did a similar code but was facing issue with the carry part.. Thanks for the help buddy>>!!
@@bishalchatterjee745 welcome bro :)
Thanks for the solution. @Amit Shukla but what would be the time complexity of this solution?
@@taranjotsingh2374 O(n)
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;
}
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
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.
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))
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)
Question 3 : easiest and shortest solution
string s1, s2;
for(int i=0;i integer to string
vector arr;
for(int i=0;i
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
(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
Can you help in explaining the time complexity of this solution
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.
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.
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
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
Nice approach 👍
@@avibirla9863Thankyou
bhai ye hi mene socha tha ,kya ye cases waki bakchodi karni😅
This line "Hello jii this is love babbar" gives us a peace and motivation that i can do 🙏🙏🙏
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());
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);
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........🌹
This placement series is awesome bhaiya and best ever series on dsa ever...
me Saturday-sunday pura din beth ke, video dekhta hu aur practice bhi karta hu,
Thank you bhai💛🤗
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
another approach to solve rotate an array
class Solution {
public:
void reverse(vector& arr, int s, int e){
while(s
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
we have to use ans.pop_back(sum);
in 3rd question to avoid reverse function
this could be also a solution of sum of two arrays.
please consider this code also
vector reverse(vectorv) {
int s=0;
int e=v.size()-1;
while(s
First
By the review of your course in UA-cam i don't visit your channel but now from this video i recommend my friend your channel.
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)
n+n kaise hoga bro ak bar explain kr de.....m confuse ho rha hu is question m
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;
}
Thank you for this placement series Bhaiya
We're learning and enjoying a lot.
q1) time complexity- O(n)
q2) O(n);
q3) O(n+m);
Best way of teaching. Quick and quirky. Please keep on making such videos.
"check if array is sorted and rotated" Sir is question me loop i=1 se i
//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;
}
#consistency op
Both your and mine let's see who will break first 😂
I sticked until last minute. I am liking it. Let's keep the josh high🔥
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❣️🙌
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)
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
Abhi aaya hu dekhne ke pehle hi bol diya done..👍👍
bool check(vector &arr){
int count =0;
for(int i=1; iarr[i] % arr.size())
count++;
}
return count
Love bhaiya content bohot tagde level ka aa raha hai 🔥🔥😍 Aise hi banate raho
(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
🔥
beautiful approach , easy to implement thanks alot keep coding sir
This is way better than uploading a 4 hr video In a day and taking break for weeks
rotate by k(without using extra space)
void rotateFull(vector& num,int s,int e){
int n=num.size();
while(s
Nice
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;
}
bhaiya apka solutions ka approach ko 100 topoo ki salamii.maja araha hai dsa karna ma ab pahla stress ata tha rona ata tha.
Q1- Time=O(n)
Space=O(n)
Q2- Time=O(n)
Space=O(1)
Q3- Time=O(n+m)
Space=O(n+m)
*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
Bhaii yhaa num se pehle ampersand kyu lgaya haii btayee ga Kya
In sorted and rotated how nums[i-1]>nums[i] if it's sorted
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
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?
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)]
bhaiya apne ye kaisa logic soch liya mujhe ye dekh ke differentiation yaad aaa gya
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;
}
My approach for Q2.
int count=0;
for(int i=0;i
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).
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.
will make it as a practice from next video
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());
}
};
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; }
}
};
bhaiya aj 3no question khud se lagaye maza he aagya. confidence next level hai ab
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
attendance lecture 21
thank you bhaiya for this amazing video
aur notes ma lagta ha apke raw video file upload ho gaye haan
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
Question bahot sahi le rahe ho bhaiya...ekdum kadak level k🔥maja aagya
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).
Going great, bhaiya!
Keep it up! We are with you!
Explanation of sum of array is awesome bhaiya best explanation I did not see anywhere,explanation like yours thanx
Thanks, bhaiya, I am continuously watching your DSA series. I am loving it so much
Ek he to dil hai , kitne bar jitoge babbar bhai
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;
}
};
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
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))
Check if rotated and sorted array - Without extra space ( temp ) --- > O(n) | O(1) solution:
class Solution {
public:
void rotate(vector& nums, int k)
{
k = k%nums.size();
int s1 = 0, e1 = nums.size()-k-1;
int s2 = nums.size()-k, e2 = nums.size()-1;
while(s1
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)
Top level consistency🔥🔥🔥
on leetcode, the first question states that you have to solve it using constant space complexity. the solution provided has linear space complexity.
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
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
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
Commenting for reach
Babbar bhai great job ! ❤️
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
1) time complexity- O(n)
2) O(n);
3) O(n+m);
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
good bro nice solution
@@prakharlowanshi1781 tq
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)
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());
}
};
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());
}
};
Request: web dev ka course kb tk aayega because dsa itna achaa h to pta nhi dev fir Kitna khaas hoga
@@shubham5934 bro jitna jldi hoga utna shi rhega
Tu 4th year mai hai?
@@shubham5934 bhia web development itna important nahi jitna dsa he. Aur waise bhi web d samajh sakte ho apne ap se par dsa samajhne ke liye teacher. So let bhaiya teach dsa first
Ruko jara sabar karo
ua-cam.com/video/tAQIKe0UGH4/v-deo.html
Semester chalrahe hai
Semester khatam hote hi Sare vedios dekhlungaa
APP DAREHAATE RAHOO
BABBAR BRO
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).
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
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
go and watch NeetCode's solution, the way that guy solved the question is exactly how you should approach and analyze a question
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
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
add 2 arrays : diffrent approach
solution :
vector findArraySum(vector&a, int n, vector&b, int m) {
int digit1=0;
for(int i=0;i
I like your course more than MIT Data Structures and Algorithms!
bhaiya teeno questions solve hogaye thankyou so much bhaiya>>>>>>