Re 2. Problems on Recursion | Strivers A2Z DSA Course

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Notes:
Understand recursion by printing something N times: takeuforward.org/recursion/in...
Print name N times using recursion: takeuforward.org/recursion/pr...
Print 1 to N using recursion: takeuforward.org/recursion/pr...
Print N to 1 using recursion: takeuforward.org/recursion/pr...
Sum of first N numbers: takeuforward.org/data-structu...
Factorial of N numbers: takeuforward.org/data-structu...
Reverse an array: takeuforward.org/data-structu...
Check if a string is palindrome or not: takeuforward.org/data-structu...
Fibonacci Number: takeuforward.org/arrays/print...
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[@takeUforward]

Please do comment how was it ? :)

[@DurgaShiva7574]

public class recursiveCode {
public static void main(String[] args) {
//print n to 1 using backtrackign, i.e without using n-1 approach
backTrack(0,4);
}
static void backTrack(int start, int n)
{
if (n == start)
{
return;
}
backTrack(start+1,n);
System.out.println(start+1);
}
}

[@sehejwahla5437]

void f(int i, int j){
if(i==-1) return;
f(i-1, j);
if(i==j) return;
cout

[@PRABHATKUMAR-ek5cu]

Thanks striver for this series, waiting for DP series

[@rishavkavlog]

As always fantastic
Soln:
#include
using namespace std;
void print(int i, int n)
{
if (i > n) return;
print(i + 1, n);
cout N;
print(1, N);
return 0;
}

[@sehejwahla5437]

Bhai thank u so much. Only coz of u i am finally understanding recursion and backtracking

[@anutoshghosh7893]

In recursion while going, things happen while in backtracking, in coming back things happen, that's why the term backtracking, so well explained!!

[@satviikk]

i missed it, thanks'

[@vaishnaviupadhyay2239]

Code for printing N to 1 using backtracking concept:
#include
using namespace std;
void print(int n , int i)
{
if(n

[@deepchangani2004]

Bro i didn't understand how this code was executed if we write i +1 then why code was not starting from 1 to n? Please give answer

[@pwnified1736]

@@deepchangani2004 bhai isten to what striver said he said ki aap print ko pehle likhne ke jagah function ko wapis call kar rhe ho toh woh jo number pehle print hona tha last mei print hoga and so on...jasa n= 4 then woh stack mei store hote jaega in order 1,2,3,4 uske baad jab base case reach hoga toh woh function call karte rahega so order will be changed to 4,3,2,1 😉

[@user-ri5gw3ty1y]

Nice answer, it does require two vars...

[@jitendrabaravkar]

​@@user-ri5gw3ty1y
#include
using namespace std;
void f(int cnt, int n) {
if (cnt == 0) return;
cout n;
cnt = n;
f(cnt, n);
return 0;
}

[@manashpratimbhowmick7211]

@@deepchangani2004 notice carefully void function k bad wo firse bar bar function ko call karrha ha jis karan use print karne ka mauka hi nahi milraha in short print line hit he nahi karraha ha and jab wo base condition ko hit karega last m tb jab wo back jayega yafir backtrack krega har function m tab wo ak ak kar k print krega .

[@akash5653]

This gave me some confidence and understanding of Recursion. Thank You so much!

[@rohitsrivastava3768]

Dude you are underrated, amazing amount of effort it takes!

[@aadityabuchale15]

He is No more UNDERRATED..!!

[@saiteja2003]

yeahhhh
@@aadityabuchale15

[@vm1662]

Can't believe all of this is for free!!! Thank you so much take U forward team!

[@deepakagrawal2392]

For the problems #3 and #4, you don't need to pass two parameters. f(i) will be sufficient!

[@nikhildotasara8788]

I think for the first one only one parameter is enough.But for the second one i have used two.

[@omkarraskar8664]

@@nikhildotasara8788 for second too you can do with 1 parameter

[@lakshmivenkatavarun8406]

@@omkarraskar8664 How bro?

[@ramanahlawat398]

@@nikhildotasara8788 u r correct bro we need another parameter in #5th problem

[@mahadevgopanpalli6194]

@@ramanahlawat398 for all problems only one parameter is sufficient bcoz if we see every prob statement clearly
Although, he was teaching , so he need to take such example

[@Ironman2.0xCode]

With each passing vdo of Striver bhaiya i saw , I am becmoing fan of him. When it comes to DSA i always recmmmend to watch Striver's vdo to my friends and juniors. The clarity in each topic is just wow.

[@kushagraahire1871]

public static void main(String[] args)
{
Scanner input = new Scanner ( System.in );
System.out.print("Enter the value of n : ");
int n = input.nextInt();
int i=1;
printNumbers(i,n);
input.close();
}
static void printNumbers(int i , int n )
{

if(i>n)
{
return;
}
else{
printNumbers(i+1,n);
System.out.println(i);

}
}
}
You are one of the best DSA tutor

[@akhileshsaga217]

Question 5 - Java Code :
public class PrintNumbersinReverseUsingBackTracking{
static void reverse(int i,int N){
if(i>N){
return ;
}
reverse(i+1,N);
System.out.println(i);
}
public static void main(String args[]){
int N=10;
reverse(1,N);
}
}

[@muhammedhuzaifa1392]

​@clawseekgaming4681N is input and compareing with so i cant exceed the limit of i=1; other wise you will face swgmental fault error 🙃

[@eshitakrishnet2291]

@@muhammedhuzaifa1392 didn't get u, can u explain further?

[@bhaswatiroy1433]

I really don't know if someone explains as good as you do !!!! 👏 Unbelievable Efforts

[@arsalaanahmed4083]

zero effort cz he is a genius.

[@shubhamagarwal1434]

#Free Education For All... # Bhishma Pitamah of DSA...You could have earned in lacs by putting it as paid couses on udamey or any other elaerning portals, but you decided to make it free...it requires a greate sacrifice and a feeling of giving back to community, there might be very few peope in world who does this..."विद्या का दान ही सर्वोत्तम दान होता है" Hats Off to you man, Salute from 10+ yrs exp guy from BLR, India.

[@abhisekmishra5245]

Now I'm gaining more confidence in recursion. Amazing explaination. 😃

[@user-ps1tn1el1v]

for N to 1
we only need one parameter
#include
using namespace std;
void print(int n){
if(n == 0){
return;
}
cout n;
print(n);
return 0;
}

[@anordinaryguy207]

backtracking se nhi hora y

[@kantiwaluma4340]

@@anordinaryguy207 ++

[@bhabanipujari1238]

Write if(n

[@ayushrawat9252]

Question 5 -> Print no from N to 1 using backtracking
#include
using namespace std;
void printName(int i , int n){
if(i>n){
return;
}
printName(i+1 ,n);
std::cout

[@lex-zt6uc]

Brother why use std:: when u have imported the whole std library itself

[@ayushrawat9252]

@@lex-zt6uc it was auto completed by the compiler so i do not erase the std

[@Akash-yr2if]

Why std::

[@ayushrawat9252]

@@Akash-yr2if if you are not using "using namespace std" then we have to use Std in every cin and cout

[@Akash-yr2if]

@@ayushrawat9252 I knw that...But even after using it in the header, Why have u used it.

[@hemalathasunnapu750]

The videos are absolutely from basic. Great Explanation

[@CodingVeda]

This video really boosted my confidence for recursion. Thank you so much

[@de_ansh]

One of the best playlist on recursion, thank you so much bhaiya.

[@timothyajani7100]

You're a teacher. I just understood what I thought was magic

[@kunalwadhai777]

Bhaiya you understand the concept of recursion in a very simple way, hats of for that and I am blessed that I can learn from such a guru.

[@aryanpinto5105]

Best video for Recursion basics on RUclips!!!

[@travelwithmightymocha]

def backtrack(i, n):
if i > n:
return
backtrack(i+1, n)
print(i)
backtrack(1, 5)

[@takeUforward]

Correct

[@farazfarooqui8656]

Best material out their ... love the way u connected/distinguished recursion from backtracking

[@siddhantdwivedi7024]

you completely nailed it bruhh!!!!!!!!!!! thanks for making such a beautiful series

[@kshitijraj1320]

Finally understood the meaning of printing after and before function call,Thank you striver 🙌🙌

[@NaturesVibee]

PRINT FOR N To 1 using Backtracking
#include
using namespace std;
void print(int i, int n){
if (n

[@photozoa]

the base condition should be if (n>3){return;}

[@anshumann]

​@@photozoaaree kya bol rhe😂

[@prathamasrani5608]

​@@photozoano...

[@yashhokte1020]

Solution of the assignment given by you : print n to 1 using Backtracking :
private static void print(int i, int n){
if(i>n)
return;
print(i+1,n);
System.out.println(i); // Backtracking !!
}
public static void main(String[] args) {
print(1,5);
}
Thankyou bhaiya for making this concept in such a crystal clear way !! Loved the way u simplify the things to us , May wish you will achieve all your sky offers to you ..

[@Iqbal00123]

You are right brother , he is great at teaching...

[@yashhokte1020]

@@Iqbal00123 yes brother

[@parthsalat]

Thank you.
I understood it completely.
May this comment make youtube feature this video to all users.

[@coding8000]

thanks!The Explanation of your is at God Level,Best on the internet and planet earth., Please keep doing such work.,Thanks you.Love from Mumbai. Thanks a lot.

[@RahulKumar-rk1tf]

Start at: 1:08
Back Tracking : 15:25

[@mdimranhussain171]

Thanks man 😄

[@devarora6995]

#include
using namespace std;
void f(int i,int N){
if(i>N)return;
f(i+1,N);
cout

[@consolecoder6724]

This is head Recursion.

[@nishantpruthi1172]

@@consolecoder6724 issi se toh krna h bro

[@consolecoder6724]

@@nishantpruthi1172 yup.

[@nishantpruthi1172]

@@mukulchopra3378 ohfo, coding seekhi ja rhi h, mukul chopra dwara

[@S__Arslaan]

is there any other approach if we take both i and n equal to n ?

[@__ngl__]

hi, thank you so much for this video!! i've always struggled with backtracking and for the first time ever, i felt like i understood it.

[@abhaytiwari6411]

Thank for making such a wonderful series.

[@Sanjeev.Network]

recurse(int N,int final){
if(final>N) return;
recurse(N,final+1);
print(final);
}

[@sarikaabarnwal]

Class A{
public void print(int n){
if(n

[@Aman-tr4bb]

the thing is you can not use n-1;

[@lakshyasadhwani9694]

this is wrong brother. this will print from 1 to n

[@TheAditya-zt9pb]

Loved the way you explained the concept of recursion and Backtracking ,i was able to get a clear picture of it and Thanks fot it

[@omgaikwad5962]

Excellent teaching! understood the concept of Recursion

[@shashanksharma7747]

public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
fun(1,n);

}
static void fun(int i,int n){
if(i>n)
return ;
fun(i+1,n);
System.out.println(i);
}

[@amanbhadani8840]

I don't think,we need to carry value of N with i everytime if base case dont need it.Rest explanation is amazing.

[@afuadajoemmanuel5735]

Thanks for all you do Striver ❤

[@Manishgupta200]

Thankyou Striver. Before wathing this video.. in my mind, I'm feared a lot of the names say Backtracking, Segmentation, Brute force approach. .. But you cleared my doubt. Not, it's easy to analyse every term

[@lathishgajula387]

just do condition(i>n) return ;
then apply function (i+1,n)
then cout

[@_vengeance]

EZ

[@gyaneshjha19]

Yaar thanks a lot striver, I was unable to understand the recursion even after doing 30 to 35 questions ,and ur this single video helped me in getting the concept easily,
After watching this video I was like my life was lie 😂

[@yashhokte1020]

bro we are sailing on the same boat , would like to discuss something related recursion to you !! shall we?

[@shrunkhalaraut7789]

Thanks for the video. 1st time I understand the backtracking.

[@FireGamerz-of6vv]

Excellant getting confident day by day

[@percussionistbypassion2931]

C++ - Print N to 1 using recursion but with backtracking:
#include
using namespace std;
int n, i = 1;
void f(int x,int y);
void f(int i,int n)
{
if(i>n) return;
f(i+1,n);
cout

[@tanveer.shaikh]

for the third question we could have a single function call with n only

[@ESaiCharanKNetha]

Just started watching your recursion playlist

[@sunnybunny4395]

if(i>n) return;
print5(i+1,n);
cout

[@zorodluffy6415]

#include
using namespace std;
void printn_1_backTracking(int n, int i) {
if (i > n) {
return;
}
printn_1_backTracking(n, i + 1);
cout

[@EjazAhmed-pf5tz]

def number(i,n):
if n

[@padmasrinivasanmanivannan]

def func(i,n):
if(n

[@aashritamutkiri5071]

class Recursion
{
public static void main(String[] args) {

int n = Integer.parseInt("5");
printReverse(5, 5);
}
static void printReverse(int n, int num)
{
if(n == 0)
return;

printReverse(n-1, num);
System.out.println(num-n+1);
}
}

[@cheemus7465]

minus use ni krskete vro

[@lakeshkumar1252]

def f(a,b):
if a>b:
return
f(a+1,b)
print(a)
f(1,3)
#thanks for the content

[@rishikeshmore5665]

void print(int i, int n){
if(i>n)return;
print(i+1,n);
cout

[@user-ti3bd8mp1w]

understoood
Thank you striver for such an amazing explanation

[@priyanshusaxena7157]

// Use backtracking to print N to 1
#include
using namespace std;
void print(int i, int N)
{
if(i > N)
{
return;
}
print(i + 1, N);
cout

[@sivaramapandiyan5117]

Question 5:- [Python Solution (BackTracking)]
def printNumber(i,n):
if(i>n):
return
printNumber(i+1,n)
print(i)
number = int(input())
printNumber(1, number)

[@cheemus7465]

oooooo

[@thebackendcoder3147]

Great video Striver
function print(i, n) {
if (i == n + 1) {
return
}
print(i + 1, n);
console.log(i);
}

[@aniketsharma472]

for Q5 in C++
#include
using namespace std;
void num(int a,int n){
if(a>n){
return;
}
num(a+1,n);
cout

[@radharamamohanakunnattur3035]

Understood!! Awesome explanation

[@pritampaul6747]

def backtrackNum1(i,n):
if i>n:
return
backtrackNum1(i+1,n)
print(i)
backtrackNum1(1,3)
Thank you so much, Sir.

[@riyarsharma4667]

Thank You Striver!!

[@lalitbisht8381]

Very nice choice of question to make the base strong❤

[@amruthaammu7463]

Wonderful series

[@user-js1rx8rs9p]

thank you striver . you made my day

[@aneezahayatshaikh3176]

Really very thankful to you sir, so well explained

[@shubhamagarwal1434]

Very nice explanation...God Bless You

[@sarangkumarsingh7901]

Understood awesome.....

[@akshay84080]

amazing playlist of dsa it's like wow

[@sakshamnda4802]

void linear(int i,int n){
if(i>n){
return;
}
linear(i+1,n);
cout

[@abhaymishra8486]

very good explanation i have cleary understood the concept of backtracking.😎

[@hckermanrober03]

vaiya totally clear the basic recursion😍

[@shreyasmekap1452]

Amazing lecture sir! Thanks a lot!!
Last qs ans =>
#include
using namespace std;
void print(int i, int n){
if(i>n) return;
print(i+1,n);
cout n;
print(1,n);
return 0;
}

[@TanishGupta-hw5gs]

In python(for python users :) ):
def gg1(n,i):
if n

[@Nishantkumar-oh9th]

For Printing the counting 2 parameters makes it complex .
void fun(int n){
if(n==0)
return;
cout

[@ganeshv045]

Understood , super explanation

[@dhirajjha29]

void linear(int i,int n){
if(i>n) return;
linear(i+1,n);
cout

[@drishtirai864]

Thank you ! Understood :))

[@shubh13799]

def pattern(i, n):
if i > n:
return
pattern(i+1, n)
print(i)
print(pattern(1,3))

[@Munchen888]

Thank you for good explanation and last hometask)

[@Mr_kp66]

Why we need two parameters to pass ?
Instead we can use only one parameter by
Void print(int i)
{
If(i

[@swarnabd202]

Python implementation of print N to 1(using backtracking):
def printNto1(lower,upper):
if(lower==upper):
print(lower)
else:
printNto1(lower+1,upper)
print(lower)
printNto1(1,7)

[@smoothierudiee]

amazing video striver!

[@stith_pragya]

Thanks Striver Brother 🙏🙏🙏

[@arnabdeb1486]

#include
using namespace std;
void backTrack(int i,int n){
if(i>n)
return;
backTrack(i+1,n);
cout

[@RohitRaj-hl6ji]

This code is printing the numbers or only i n times??

[@jasmeetsingh5174]

public class sec {
static void f(int n,int i){
if (i>n) return;
f(n,i+1);
System.out.println(i);
}
public static void main(String[] args) {
int n=50;
f(n,1);
}
}
java approach for printing n to 1 using backtrack (thanks striver bro)

[@rishabhverma7329]

Thanks for the great content

[@dipannita7436]

helpful video. Thanks a lot.

[@VikashKumar0409]

def f(i,n):
if(i>n):
return
f(i+1,n)
print(i)
n=int(input("enter the number"))
f(1,n)

[@abdoo7197]

what a great explanation

[@ravalikatalks5285]

thanks alot bhaiya for the best explanation!
😊

[@NazeerBashaShaik]

Understood thank you.

[@ashanagar5226]

Thankyou so much just amazing explanation

[@jayesh8401]

Well explained 🔥

[@CodewithKing360]

void print(i,N)
{
if (i>N) return;
print(i+1,N);
cout

[@anandkishor5737]

In Python:
def recurse(i,n):
if i > n: return
recurse(i+1,n)
print(i)
recurse(1,5)

[@chitrankusarkar7278]

20:52
ques 5
void printNToOne(int i, int n)
{
if(i > n)
return;
printNToOne(i + 1, n);
cout

[@sakshi-ok8zu]

recursion (2/21) ✅
void print(int i, int n) {
if(i > n) return;
print(i+1, n);
cout

[@raghavsinghal22]

Striver's backtracking understood!