Sir I was struggling to get some confidence to solve a new recursive problems and I literally watched many videos on recursion but never got the intuition but when I watched your recursion and dp... Playlist I am able to get the intuition for any new recursive problem. I can't explain how much gratitude I've for you but still thanks alot Sir :) Really admire your hard work

best programming content on youtube or on any platform.. nobody can match u... bhai recursion pdha rhe ho..next backtracking pdha dena..maza aa jaega

I can say this with utmost confidence those insignificant amount of dislikes are from people who have not watched and performed previous videos of this playlist, otherwise this is the GOAT recursion course!

Aditya, this is some different level of coding and so easy to understand especially for confusing topic like Recursion. First time I understand why people say to have "leap of faith" while solving recursion... Kudos to your efforts

Bhai adiya verma, bhagwan tujhe lambi umer de. first time aise vedio mili hai jisme itna sahi btaya hai

both striver bhai and aditya bhai is dsa guru but guys seriously aditya bhai recursive playlist is more better than striver,btw striver recursion playlist also too good but aditya bhai recursion playist is just love....

As you said in first video... Writing the code was really a cakewalk ! ❤️❤️

Bro please make BACKTRACKING videos.Your explanations are unanimously best on KZfaq.Thanks a lot man❤️

You are doing great please go forward so that we can get more benefits and thanks for sharing your knowledge freely 🙌🙌😁

I could not find any other tutorial explaining intuition behind recursive tree solutions. This is the the only one. Thanks a ton!

Watched all videos of every playlist Till now and waiting for more.. Thankyou for making such an important channel

Thanks to humme bolna chahiye __/\__ I have had difficulty in understanding recursion but thanks to you, I am improving with every day.

Broooo the way you explain the solution is something else🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥

A very happy teachers day to the best teacher Aditya sir..you are the best teacher I ever had..you made recursion so easy I can't even thought.hope the almighty bless you with lots of happines and growth..😊😊

Java Code private static void permutationFind(String input, String output) { if(input.length()==0){ System.out.println(output); return; } String out1 = output; String out2 = output; out2 += input.charAt(0); input = input.substring(1); permutationFind(input,out1); permutationFind(input,out2); }

Sachme vaiya apke vdo ki starting k music sei ek motivation milti he❤️

code for c++ vectorm; void solve(vector nums,vector op, vector &m) // m ko by address paas kiye as jo bhi changes m me ho wo vector wale fn me reflect ho and nums aur op ko pass by value kiye uska changes vector fn me reflect nahi hoga { if(nums.size()==0) { m.push_back(op); return; } vectorop1=op; vectorop2=op; op2.push_back(nums[0]); nums.erase(nums.begin()+0); solve(nums,op1, m); solve(nums,op2, m); } vector subsets(vector& nums) { vector m; vectorop; solve(nums,op,m); return m; }

Python 3 Code Time Complexity - O(2**N) Space Complexity - O(1) [Excluded Auxilary Stack Space] Code: def generateSubsets(Input): def helper(Op, In): if(In == ""): print(Op) return Op2 = Op Op2 += In[0] In = In[1:] helper(Op, In) helper(Op2, In) helper("", Input) Input = "abc" generateSubsets(Input)

Once again thanks Again. did saw the code in video just watched the tree part and design the code by myself. and did little tweak to avoid extra variable. and here it like you said cakewalk. thanks Aditya. posting the code also in java. static void subSet(String op, String ip) { if (ip.length() == 0) { System.out.println(op); return; } subSet(op, ip.substring(1, ip.length())); subSet(op + ip.substring(0, 1), ip.substring(1, ip.length())); }

when he said "The guy who made dp/recursion easy". He really ment it!!!

Did myself in 15 min! thanks bro!

Best teacher for coding

@Aditya Verma, Videos are very interesting and never forgot your teaching style... Thanks a lot keep going... Please upload some graph related videos.

Best Print all subset solution till now ❤️

at end of the video also mention time and space complexity..it will be useful for all of us.thank u

recursion jaadu hai bhai .. Thanks a lot brother! you are the best! Please add graphs playlist as well

Please do a video on: Print all permutations of a string!

Glad to see your lecture brother👌 Take love from Bangladesh ❤🇧🇩

Can you please tell us the order of the upcoming topics like will you be teaching backtracking or graphs or trees. It will be helpful for us and will give us clarity.

this is the best explanation I have ever seen!

Please Next Question -> Phone Keypad Combinations print.

@Aditya verma thanks a lot for removing the fear of recursion !! Before your videos, I used to avoid recursion wherever possible. Can you please suggest something on where to use the return statement generally in any question.

bhaiya please add questions on binary trees which is totally dependent on recursion. Thanks

Problem: Product of Array Except Self btw, Great content, thanks! :)

bhai feel dila de recursion ke , thanks a lot bhai

Most liked youtuber of mine❤️

Keep the good work going ! Amazing videos. 👍😃

bas ek step tak chalakar dekh lenge IP-OP method main bhi? fir use hisab se likh denge code, baki apne aap kam kar jaega! pls correct me if I am wrong! Awesome way to explain man!

Hi brother, make a video on proper use of returns in recursion, most of the time I am out of clue when it comes to returns statement usage. That will be a great help. Thank you

Thanks for the detailed explanation. Plesse could you make us undersatntd that what is the importance of the "return " statement and how missing it will cause a segmentation fault .Thanks

Thanks for such detailed explanation. One small suggestion from my side : "Try Using Dark theme Mode" a lot of the students like me watch your videos at night, and dark theme will be easy on the eyes. Thanks you once again you're doing an amazing work :)

/* * Subarray: * 1: Contiguous * 2: Order of appearence is maintained * 3: Every subarray is a subset. * 4: Every subarray is a subsequence * 5:For an array/string of size n, we can have ( (n*(n+1))/2 + 1 ) Subarray in * total i.e (n*(n+1))/2 non-empty subarray + 1 Empty subarray. * * Subsequence: * 1: May or may not be contiguous * 2: Order of appearence is maintained. * 3: Derived by deleting some or none of the elements of the sequence without * changing the order of * the remaining elements. * 4: Every subsequence is subset. * 5: For a sequence/array/string of size n, we can have 2^n Subsequence in * total i.e ((2^n)-1) non-empty sub-sequences + 1 Empty Subsequence. * * Subset: * 1 : May or maynot be contiguous * 2: Order of appearence may or may not be maintained. * 3: Subsets and subsequence ae exactly same thing the only difference between * them is that subsequence should maintain order of appearence of elements * while subsets don't need to preserve order of appearence of elements * for ex : {1,2} and {2,1} represent the same subset however while * writing/listing the subsets of the sets you can either write {1,2} or {2,1} * it doesnot matter but you can not list both at the same time * since they are exactly the same thing. * * Ex: arr[]={1, 2, 3} * * Subarrays : * * [1] * [1,2] * [1, 2, 3] * [2] * [2,3] * [3] * * * Subsequences: * * [] * [1] * [1, 2] * [1, 3] * [1, 2, 3] * [2] * [2, 3] * [3] * * Subsets: * * [] * [1] * [1, 2] or [2, 1] * [1, 3] or [3, 1] * [1, 2, 3] or [1, 3, 2] or [2, 3, 1] or [2, 1, 3] or [3, 1, 2] or [3, 2, 1] * [2] * [2, 3] or [3, 2] * [3] */ //Iterative Approach //Method 1 : /* For arr[] = {1, 2, 3}: Intially: subsets :[[]]. (Add an empty subset) Iteration 1: (arr[0]=1) Pair 1 with all existing subsets i.e [1] subsets: [[],[1]] Iteration 2: (arr[1]=2) Pair 2 with all existing subsets i.e [2] and [1, 2] subsets: [[], [1], [2], [1, 2]]. Iteration 3: (arr[2]=3) Pair 3 with all existing subsets i.e [3], [1, 3] , [2, 3] and [1, 2, 3] subsets: [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]. */ class Solution { public List subsets(int[] arr) { List subsets = new ArrayList(); subsets.add(new ArrayList()); // Start with the empty subset for (int val : arr) { int size = subsets.size(); for (int i = 0; i < size; i++) { // Create a new subset from the existing subsets by adding the current number to // it List subset = new ArrayList(subsets.get(i)); subset.add(val); subsets.add(subset); } } return subsets; } } // Method 2 : Bit Manipulation /* * For example : arr[] = {1, 2, 3, 4}; there are 16 (i.e 2^4) sub-sequences. * If we carefully notice it is nothing but binary numbers from 0 to 15 which * can be shown as below: * * 0000 {} * * 0001 {1} * * 0010 {2} * * 0011 {1, 2} * * 0100 {3} * * 0101 {1, 3} * * 0110 {2, 3} * * 0111 {1, 2, 3} * * 1000 {4} * * 1001 {1, 4} * * 1010 {1, 4} * * 1011 {1, 2, 4} * * 1100 {3, 4} * * 1101 {1, 3, 4} * * 1110 {2, 3, 4} * * 1111 {1, 2, 3, 4} * * * Starting from right,whenever we encounter set bit(1) in a number at ith * position the we print the element present at (i-1)th index in the array. */ // Intuitive Implementation class Solution { public List subsets(int[] arr) { List subsets = new ArrayList(); // Total number of subsets is 2^n int totalSubsets = 1

Sir Your Video are really worthy and denote your true hardwork behind it. But Please try to release your series in an order from easy to hard. Do number them if necessary.

bhai mai normally comment nahi karta but thanks bro i will surely support you on pateron when i'm able to make enough money to support you

zindagi mein pheli bar smjh mein aaya hai

instead of erase function we can just pass ip.substr(1); it returns the string remaining after the first element

Hare Krishna!🙏🙏 Great Work

for leetcode question [aditya verma method], array as input void subset1(vector ip, vector op,vector& ans){ if(ip.size()==0){ ans.push_back(op); return; } vector output1=op; vector output2=op; output2.push_back(ip[0]); ip.erase(ip.begin()+0); subset1(ip,output1,ans); subset1(ip,output2,ans); } vector subsets(vector& nums) { vector op={}; vector ans; subset1(nums, op, ans); return ans; }

Python Code (LC Subsets) def subsets(A, output, result): if len(A) == 0: output.append(result) return L = result[:] R = result[:] R.append(A[0]) A = A[1:] subsets(A, output, L) subsets(A, output, R) A = [1, 2, 3] output = [] result = [] subsets(A, output, result) print(output)

Thank you so much for this great playlist.

time complexity is o(2^n) .am i right ??

Thanks for making recursion so easy!

Wonderfully explained!! thank you.

Bhaut mast smjhaya bhai!!

is can we able to do with array because we are not able to small the input with array?

Bhai aapka Flipkart and Nutanix interview experience share kab karoge.

Thank you Bhai, bohot hi easy understanding k sath problem samjadi aapne...! keep it up! Here is my python code def subsets(self, nums: List[int]) -> List[List[int]]: self.res = [] op = [] ip = nums self.solve(ip, op) return self.res def solve(self, ip, op): if (not ip): self.res.append(op) return op1 = op[:] op2 = op[:] op2.append(ip[0]) ip = ip[1:] self.solve(ip, op1) self.solve(ip, op2)

Way better Sound quality in this one.

class Solution { private: // To solve this first draw Recursive tree then code it void solve(vector& ans,int index,vector nums,vector output) { // base case if(index>=nums.size()) { ans.push_back(output); //before returning we have to push the output in ans return; } /* note pahle exclude ka recursion call krenge uske baad include krenge elements ko nahi toh pahle hi elements include ho jayenge toh ans me original array hi print hote rahega */ // exclude that element and Recursion call solve(ans,index+1,nums,output); // include that element int element=nums[index]; output.push_back(element); // Recursion call => Includded R.C solve(ans,index+1,nums,output); } public: vector subsets(vector& nums) { vector ans; vector output; int index=0; solve(ans,index,nums,output); // we create this function for recursive call bcz we have to pass each time (ans,index,nums,outpu)t and not returning anything only have to update the ans; return ans; } }; // thanks me later

java - solution public class PowerSet { public static void main(String[] args) { String ip = "abc"; String op = ""; subset(ip,op); } private static void subset(String ip, String op) { if(ip.length()==0) { System.out.println(op); return; } String op1 = op; String op2 = op +ip.charAt(0); ip = ip.substring(1); subset(ip, op1); subset(ip,op2); } }

Thanks ! Awesome Explanation...

thanks a lot for explaining so beautifully

//Java Recursive functiion of this code static void solve(String input,String output){ if(input.length() == 0){ System.out.println(output); return; } //now take 1st char of input string char ch = input.charAt(0); //take char at 0th index of input string solve(input.substring(1),output + ch); //ignore char at 0th index of input string solve(input.substring(1),output ); }

bhaiya please make video on Asymptotic notations and time complexity and space complexity

Please DP ki playlist pure Complete kardo.. LIS, Kadanes's wagera bache hain.. Plz next usse complete kardo..

thank you.

bhaiya can you provide us with a list of practice questions of about 20-30 questions,it would be very much helpful ..as there are thousand of questions there on the internet

Python Implementation........ def subsets(self, nums: List[int]) -> List[List[int]]: output = [] def helper_fun(start,temp,output): output.append(temp) for i in range(start,len(nums)): helper_fun(i+1,temp + [nums[i]],output) helper_fun(0, [],output) return output

the same can be done with vector right? means to get all subsets of int array

bhaiya, kya aap Graph series bhi upload karoge? aapka padhaya hua bohot acche se samajh mein aata hai, pls graphs bhi upload kardena

awesome explaination bro. thanks

#include using namespace std; void subsets(string input,string output){ if(input.size()==0){ cout

Okayyy.. So, Smaller input will be same for all common nodes of a common generation? Hona vi chahiye... its so obvious .. haina

Great teacher

HEY, thank u for such a great explanation, only one doubt, can't we keep output as one variable instead of taking two different op1 and op2?

It does not work for array like ip: [1,2] and op: [[], [1], [2], [1,2]]

Thanks .. great explanation

how to memoize this. I am getting TLE on large test cases.

If we call recursive function first on op1 and then op2 , it gives error . Why??

how would you deal with negative number? it's taking -sign as a character...

Bro backtracking aaegi? Yaa nahi? Please bata do

Recursion OP!

bhai ap kahi bhagwan nhi ho n ,glti se prithvi pe aa gye ho , kyoki itne ache se insan to sikha nhi skta hai

@Aditiya how is input/output method different than choice diagram(dp)? In both we either choose to have it in the output of not... Is it just that one has optimal substructure and other has not?

Aditya bhaiya number repeat ho sakte hai tab kese karenge?? October leetcode challenge ka second question hai....

bhaiya you are best

You are doing great work. Can you pls take combination sum problem from leetcode as well.

Jab jab input khali ho raha hai ..tab tab hame output mil raha hai , But base condition runs only once isn't it ? I am confused asf.

What is the time complexity of above solution?

Solution of subsets problem in javascript:- function subsets(input,output){ if(input.length

Can we remove the last index instead of the first one ? if we remove last index then the time taken to shift the array is 0 essentially because we are just stripping of the last value.

Bhai array pe DP kaise lagaye,uspe bhi vedio banao..Thanks in Advance.

Extremely thanks

Bhaiya subset wala selection problem nhi hga kyuki hum select krte h please reply..

Sir the time complexity of this code will be 2^n??

Hi your patreon page is not supporting any debit card or PayPal and I have numerous times .plz use join of KZfaq instead to share your notes. THANKS BTW loving your videos.

bhai bhut pyaara

Bro,please please make a series on sliding window technique problems @Aditya Verma

Thanks for this video

i have done it using only IBH method . So my question is why do we should learn this method ?