Print Subsets | Print PowerSets | Print all Subsequences 

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

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....

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

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!

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

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

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

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

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

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

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

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..😊😊

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; }

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

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); }

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())); }

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

/* * 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

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

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

Did myself in 15 min! thanks bro!

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.

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

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

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

@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.

Keep the good work going ! Amazing videos. 👍😃

//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 ); }

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

Please Next Question -> Phone Keypad Combinations print.

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; }

this is the best explanation I have ever seen!

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

Best teacher for coding

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

Best Print all subset solution till now ❤️

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

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

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!

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

bhai feel dila de recursion ke , thanks a lot bhai

Most liked youtuber of mine❤️

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)

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

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.

Thank you so much for this great playlist.

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

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 :)

Thanks for making recursion so easy!

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

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

Hare Krishna!🙏🙏 Great Work

Wonderfully explained!! thank you.

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

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

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

Thanks ! Awesome Explanation...

Bhai aapka Flipkart and Nutanix interview experience share kab karoge.

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

Way better Sound quality in this one.

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

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); } }

zindagi mein pheli bar smjh mein aaya hai

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

thanks a lot for explaining so beautifully

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

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

Bhaut mast smjhaya bhai!!

What is the time complexity of above solution?

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

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

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

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

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

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)

awesome explaination bro. thanks

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

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

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.

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

please put atleast subtitles sir, it will be helpful for us

Thanks .. great explanation

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

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?

@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?

Recursion OP!

bhaiya you are best

Extremely thanks

I tried this approach by taking String input = "abc" this is working fine.. but i tried same in int array [1,2,3] then it is not giving correct output ( tried in java ).. can anyone help me please .. ? this is my code public static void solve(ArrayList ip,ArrayList op ){ if(ip.size()==0){ System.out.println(op); return; } ArrayList op1 = op; ArrayList op2 = op; op2.add(ip.get(0)); ip.remove(0); solve(ip,op1); solve(ip,op2); }

Time complexity will be: (2^(N+1)) - 1

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.

ye wala code run to hora hai but test cases pass nhi krr rhaa koi bhi

void solve(string s, string op, vector &v){ if(s.length()==0){ v.push_back(op); return; } string op1=op; string op2=op; op2.push_back(s[0]); s.erase(s.begin()+0); solve(s,op1,v); solve(s,op2,v); return; } vector powerSet(string s) { //Your code here string op=""; vector v; solve(s,op,v); return v; }

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

why we are not using & in input and output

Thanks for this video