Solve any Pattern Question - Trick Explained | 22 Patterns in 1 Shot | Strivers A2Z DSA Course

[@takeUforward]

Let's march ahead, and create an unmatchable DSA course! ❤ Do consider to give us a like, and a subscribe, means a world to us..

[@giit85]

❤️

[@bhanupratapsharma6158]

competitive programming content plz sir how to reach candidate master i was stucked on specialist

[@tuhindas9777]

Thank you dada love from Kolkata

[@23cash86]

STL next?

[@techyouknow8026]

Happy new Year , Please Sir upload all the basics lecture within 2 months as I trust you so I can request you please upload frequency maintain , I like all your video after watching it completely, also you are helping others to achieve their new year resolution.

[@sachinboreoffl]

Tip: When you're printing spaces in a pattern, replace the space with any other symbol like this "-", it helps in understanding what we are doing.

[@RahulChacko]

❤

[@jaydeeppatil7936]

Thanks a lot it really helped me

[@anusha6033]

nice one thanks for sharing

[@suryasaipalthi]

thanks for sharing that did change my way of thinking somewhat

[@harshyadav1214]

very good approach

[@sumitapathak2900]

I was losing confidence in my DSA skills after being away for so long. This video sharpened my thinking as well as brought back that confidence. Thanks a lot !!

[@FRACTALUMAR]

did you solve them by your own or first watched ,then, practice them ....

[@vahramirpunit]

@@FRACTALUMAR i first watched whole video once, then i solved most of the problems on my own, if needed i came back to video.

[@muhsin.k]

Intro: 0:00 BuildUp Logical Thinking w/ patterns: 0:20 Pattern_01: 1:34 4 Rules for solving patterns: 2:24 Pattern_02: 14:17 Pattern_03: 17:33 Pattern_04: 19:53 Pattern_05: 21:00 Pattern_06: 23:39 Pattern_07: 24:33 Pattern_08: 31:10 Pattern_09: 34:16 Pattern_10: 35:12 Pattern_11: 39:22 Pattern_12: 42:15 Pattern_13: 46:52 Pattern_14: 48:42 Pattern_15: 50:57 Pattern_16: 53:03 Pattern_17: 54:42 Pattern_18: 59:45 Pattern_19: 1:01:28 Pattern_20: 1:06:36 Pattern_21: 1:12:29 Pattern_22: 1:15:41 Like!!: 1:21:17

[@takeUforward]

Thankyou 🙏🏻 added

[@desimemerrudhra8394]

@@takeUforward where r u bro, new lectures 🤔🤔😭😭

[@eleveneleven4195]

for code 14 we can also try :🙂 for(int i=1;i

[@dllychoudhary7586]

I literally did the same... because there was error that character can not add with int..so i did with your method

[@honourable8816]

​@@dllychoudhary7586 You can use cout

[@TiyaBhatt4-de8sq]

@@dllychoudhary7586 I am facing the same error in question 16 . What should I do?

[@yashiktaneja673]

​@@TiyaBhatt4-de8sq char ch='A'; for(int i=1; i

[@rajeshvarma2162]

In Pattern18, we should not hardcode the value 'E', It should be dynamic. void pattern_18(int n) { for (int i=0; i

[@Hades0210]

Exactly

[@ictfan-ly8gh]

Yes bro u are 👍

[@ictfan-ly8gh]

He could have use Char ch= 64+n

[@kartikgautam6769]

its wrong bro...The correct solution is.... for(int i=0;i

[@ImaamJIJ]

we can use this for coding ninjas practice for (int i = 1; i ='A'+(n-i); ch--) { System.out.print(ch+" "); } System.out.println(); }

[@nisha1841]

Bhaiya learning DSA from your AtoZ sheet . i am in final year I can't thank you enough for bring light in my dark future..when everyone is saying that "tumse an ho payega "I finally decided that "ab mujhse hi ho payega" thank you bhaiya .and please agle 3 month take intermediate level ka DSA sikha do.

[@AryanKumar-qj9ud]

Same stry wish the best

[@uniquescience7047]

All the best to you😁

[@known878]

Bhai tumne bola ki 3 month takk intermediate DSA sikha do aur tumhare comment ke bhi 3 months ho gaye to sikh liye kya intermediate wala?

[@nisha1841]

@@known878 ha bhai intermediate level ka almost complete hai . Strive ki AtoZ sheet se padha hai maine usme se question bnao shi rhega

[@prajjwal5520]

placement lg gyi?

[@md.ualiurrahmanrahat2400]

saw a lot of videos before to understand pattern 22. But never got it. The way you solved it!!! What can I say. We owe you big time brother.

[@chiragPatel22c]

last pattern is mind blowing!!! hard to get that logic at first sight.

[@dhruvdonsahu9972]

i am still confused

[@raavanan00792]

Thanks for the great content ❤ @tip for the pattern 18 the logic in the video is hardcoded for 'E' letter alone . Kindly utilize the below solution. void alphaTriangle(int n) { for(int i=0;i

[@samuelfernandes3301]

striver i think you made a little mistake in pattern18 your considering j as E, but the start letter depends on number of rows, therefore the code will work for more test cases void pattern18(int n){ for (int i = 1; i

[@ankushhv6972]

Hey Striver, i think for the pattern 18, we should not directly start from 'E', I think according to pattern we should be starting form ('A' + n - i + 1) until ('A' + n), for(int i = 0; i < n; i++) { for(char ch = 'A' + (n - i - 1); ch < 'A' + n; ch++) { cout

[@shaharslan3372]

Yes, Ankush I also noticed this. Using this approach makes it dynamic for all inputs of 'n'. Good Job

[@soubhikghosh6564]

It doesnot come interviews i agree. But still i would urge all to do this exercise... It will help all Software engineers or aspirants to improve their problem solving skills by seeing patterns

[@sasmitsirroco]

For pattern 7, why print spaces after astrix? Simply cout

[@nehathakur40]

you are an inspiration for programming faternity! loved the way you teach. The ease with which you have taught recursion and dp is commendable. i get to learn every day from you. Thank you for being light in lives of programmers.

[@raunak276]

fr

[@sainaid]

no doubt striver bhaiya has made this playlist the best way.

[@mehtabmultani3194]

39:22 tip for pattern 11: if the sum of every single (row+column) is 'even' then have to print 1 else 0 if((i + j) % 2 === 0) print(1) else print(0)

[@satyam-seth]

void printPattern(int n) { for (int i = 0; i < n; i++) { for (int j = 0; j

[@012srishtisaini7]

@@satyam-seth thankss.. i was having a bad time because of this

[@Cvx_ishere]

Literally, breaking whole concepts into 3 steps really really help me as beginner to understand patterns. Thankyou striver b :)

[@shuskane]

UNDERSTOOD 👍 👍👍 👍👍👍 👍👍👍👍 👍👍👍👍👍

[@kshitijsingh1990]

haha smart!

[@zakirabhabhi3175]

let n=5 for (let i=0;i

[@udaykumar-jj1gw]

@@zakirabhabhi3175 😂 😂😂 😂😂😂 😂😂😂😂 😂😂😂😂😂

[@debashmansarbadhikari6196]

For pattern number 12 we could also have done the following : int space = 2 * ( n - i )...ofcourse this variable would have to be defined within the scope of the outer loop. Besides, the problem given on the sheet and the problem redirected to by the link are actually slightly different. Here's the code snippet for solving the problem redirected to by the link : void printTriangle(int n) { for(int i = 1; i

[@abhisheksantoshbhambe7617]

thanks dude!! i thought y im not getting it, and u coreected me

[@prathameshdhanajigandule4299]

The better and correct code for pattern no.18 1:01:00 would be : void pattern18(int n){ for(int i = 0 ; i < n ; i++){ for(int j = i ; j >= 0 ; j--) cout

[@keisergg8701]

yep! I did same as you, here is my java code if anyone cares public static void pattern18(int n) { for (int i = 0; i < n; i++) { char ch = 'A'; char chs = (char)(ch + (n-i-1)); for(int j=0; j

[@prathameshdhanajigandule4299]

@@keisergg8701 Yes bro it's correct I care

[@vikasverma5685]

​​@@keisergg8701 public static void pattern18(int n){ for(int i = 0; i < n; i++){ for(char ch = (char)('E'- i); ch

[@sabashh8208]

For pattern18 if we don't want to hardcode 'E' and use it for other row numbers then we can make small change as char initial = (char) ('A'+ (n-1)); for (int i = 0; i < n; i++) { for (char c = (char) (initial - i); c

[@NikitaNair]

I was looking for this!!!

[@anishchhabra6085]

Python code for the same: for i in range(n): char = 65 for j in range(i+1): print(chr((n-i-1)+char),end="") char += 1 print(' ',end="")

[@lekhanachidananda]

Thank you Striver! I really love your videos. Small suggestion: you could have calculated time and space complexity for all the patterns.

[@sathvikvutukuri9179]

I think most of them took O(n^2)

[@Manish-wk8ni]

Thank You, we are all hooked up. This is gonna be a Great Series.

[@ritusingh8616]

Thank you for this content! This video helped a lot! Kindly correct the code for pattern 18 to adapt to different values of input... void print18(int n){ for(int i=1; i

[@tusharkumar7963]

The best video on solving patterns on KZbin! Can't thank you enough!!!😀😀

[@shaharslan3372]

Edited : Saw you correct it in the next question :-D For pattern 19 , I don't think we should hard code the number of spaces as 8 and then keep on decreasing the number by 2 with each iteration , as it will only work for input n = 5. Rather there should be a general formula : for(int i = n ; i >=1; i--){ for(int j =1; j

[@AdityaVerma-hh1ox]

This is wrong bro

[@SpewFactsBeTrueToYourselves]

i did pattern 19 myself by running 6 for loops and without seeing strivers explanation based on previous patterns knowledge , it took me 2 hrs to first draw it on book and break it into smaller chunks and then write code, row number is the game changer in second half rhombus

[@damolaOnikoyi]

Thank you Striver, preciate ya!!. I just completed all the pattern problems. The last one was quite tricky as I did not quickly understand your solution. However, I have provided my solution below for others that might find it helpful and maybe easier to understand. Intuition: Fill the 2D array (named result) with the values (n, n-1, n-2 ...) then just simply print it out (row x column)👍🏾. You can fill the result array like this : 1. Outer loop fills the layer (A layer is a square ring of numbers). 2. Get the number to be filled per layer => int num = n - layer. Hence if n = 3, layer = 0 (first one) -> n , n - 1, n - 2 ... so we print 3, 2, & 1 per square ring. 3. Middle loop iterates over the rows of the current layer. 4. Inner loop iterates over the columns of the current layer. static void pattern22() { int size = 2 * n - 1; int[][] result = new int[size][size]; for (int layer = 0; layer < n; ++layer) { int num = n - layer; for (int i = layer; i < size - layer; ++i) { for (int j = layer; j < size - layer; ++j) { result[i][j] = num; } } } for (int[] row : result) { for (int col : row) { System.out.print(col + " "); } System.out.println(); } }

[@ramfattah211]

The last pattern took me some time but I just figured it out and i'm so happy 🧠🤯. Such a great feeling😊. Thank you Striver for this amazing content.

[@abhisheksahu3453]

for pattern 15 even more simpler code : void printPattern(int n) { int i; for(i=0;i

[@hasanuzzamananuz8270]

More than the premium course I have learned a lot. The lecture delivery, explanation, and expression are great, and that makes me concentrate more to watch the video. I have done all the problems on my own using your 4 pattern-solving rules. Hats off sir ❤❤❤❤❤❤❤❤❤❤

[@zenmonk29]

today i lost my phone. it was 19 days old. went to police station to file a report, the inspector had the same name as you

[@hasanuzzamananuz8270]

@@zenmonk29 So what, Multiple people have under the same name.

[@vidhveshchangalasetty]

18th pattern We should define a const char starting alphabet of each line by taking the char ch='A'+n-i (before starting the outer for loop) void alphaTriangle(int n) { char ch = 'A' + n - 1; for (int i = 0; i < n; i++) { for (char c = ch; c >= ch - i; c--) { cout

[@karthikeyabollu7121]

pattern 20 is same as pattern 19 but just interchanging the spaces and stars variables. that means no.of stars in pattern 20 will be same as no.of spaces in pattern 19 and vice-versa.

[@takeUforward]

Good observation

[@jayeshb9323]

Absolutely loved the video..the way you explained it made it very easy to understand complicated problems

[@gunasekarb3100]

whoever found the logic for the pattern 22 must be an alien.

[@winter_hope_]

🤣

[@LUchiha-dt6xv]

I did found the logic myself(different but same answer though) but the logic used here is damn crazy🥲

[@gunasekarb3100]

@@LUchiha-dt6xv did you use spiral way?

[@LUchiha-dt6xv]

@@gunasekarb3100 Idk about that😅, but divided the whole patter into 4 parts & designed the parts & framed them one by one(basically the brute force approach u could say)

[@abhishekpattanayak1443]

@@LUchiha-dt6xv you can also try using chebyshev distance for each (i,j) from the middle-most index

[@PrabhjotSingh-o4s]

If You Start From Loop from 0 the formula will be lil bit easier helps more to understand for a beginner 23:18 For Printing ***** **** *** ** * for(int i=0;i

[@neoteric_nk]

Nice explanation sir! Just wanna give a general code for Pattern 18. void pattern18(int n){ char ch= 'A'+n-1; for(int i=0; i

[@keerthanasathish3684]

Alternate code for pattern 5 has runtime 17ms void seeding(int n) { for(int i=0;i

[@k-gos]

What will be the frequency of lectures and will you teach the complete DSA in this course? Also what will be the tentative duration of this course? Great lec btw

[@shivendradwivedi8887]

In pattern 20 , for stars and spaces we can do this way also int stars = i; int spaces = 2*n - 2*i; if(i > n){ stars = 2*n - i; space = 2*i - 2*n; }

[@spdks3485]

HAPPY NEW YEAR SIR HAVE A GREAT YEAR AHEAD AND WE ARE VERY GLAD TO GET THE HELPING PERSON LIKE YOU FOR BEGINERS LIKE US.

[@anil-chaturvedi]

Thank you for this logic-building lecture. I'm feeling confident about pattern questions.

[@detaqwer]

Hey Raj, for pattern 11 we can try this code as well which is simpler void print11(int n) { for (int i = 1; i

[@chethanprabhu4475]

Can simplify even further void print11(int n) { for (int i = 1; i

[@blockyquasar]

@@sagarshah5341 kaise bhai? same hi hai dono me, n square hai dono me

[@sagarshah5341]

@@blockyquasar I do not remember why I said that!

[@nexxerjm348]

Patter seventeen :@58:53 Solution is very Here is approach first print space then print ascending order charcater till i Then print descending order till i-1 First loop for space: j=1 to j

[@mrnonsenseju]

New exciting videos in new year... Thanks Striver Bhaiya.... Love from Jalpaiguri

[@swapnilkhandekar4157]

for gfg 18 pattern another type void printTriangle(int n) { for(int i = 0 ; i < n ; i++){ int alpha = 26 - n; char ch = 'Z' - alpha; for(int j = 0 ; j

[@UECHarshaVardhanRavipalli]

Done these all max by my own... feels confident enough to go ahead!! TQ Bhai!!😇

[@Seraexplores]

Pattern 19 simpler: def pattern19(n): for i in range(1, 2*n + 1): stars = n-i+1 if i > n: stars = i-n spaces = 2*n - 2*stars for j in range(stars): print("*", end="") for k in range(spaces): print(" ", end="") for j in range(stars): print("*", end="") print()

[@Narasimha-ms2up]

Really amazing teaching after listening 8th to 11th pattern , without any error i do 12th by my own❤❤❤

[@mahirshah763]

In Pattern 22, if someone is coding in c language, there is no min() function so it can be written like this by creating a function: #include int minimum(int a,int b, int c, int d); int main() { int num=4; int n=2*num-1; for(int i=0;i

[@sukhpreetsingh5200]

Revision done😄

[@mr.batman2348]

hello paji koi dc ya linkedin profile wagera

[@amit1305]

thanks for the hardwork man i completely understood pattern now

[@HemasaicharanK]

Thank you for your clear and concise explanation. I used to follow you for so long on LinkedIn, but didn't get a chance to watch your content. Now is the time for me to get prepared for a DSA interview.

[@rishabh_pant]

Source code ? I have learned Star space star patterns with your trick ❤️‍🔥. It would be better if you normalize and use loop from 0 or 1 for every problem :) cheers !!!

[@sagarshah5341]

Hey Striver do you think this is a better solution for Pattern 07, Pattern 08: This is in Python. But the logic can be replicated to any other language. Pattern 07: n = temp = "" content = (n*2)-1 count = 0 for i in range(1,n+1): count = (i*2)-1 spaces = int((content-count)/2) temp = (" " * spaces) + ("*"*count) print(temp) Pattern 08: n = temp = '' content = (n*2)-1 count = 0 for i in range(n,0,-1): count = (i*2)-1 spaces = int((content-count)/2) temp = (" " * spaces) + ("*"*count) print(temp) Please correct me if I am wrong. Just a newbie.

[@luckshaeey]

last pattern was awesome. I followed another method and figured out some another pattern and symmetricity in it. But your approach was very informative too.

[@darkexodus6404]

Same. From outermost square to the innermost square. I did it like this.

[@Jaishreeram-ly2jm]

can u share your method?

[@saisameer780]

Thank you GOAT !!! Never understood patterns this good .

[@thewave1996]

Thanks bro, you are giving your best to teach us and i will assure u that i will also give my 100% so that your efforts will not be a vain. I will complete this course and solve your 180 questions so that that i can place in a good company. once again thank u so much for your in depth assistance. 🙌

[@user-zy3mz6ml7c]

You're answer for question 18 only gets correct answer when n is 5 because you hardcoded E into the for loop. The answer should be: void print18(int n){ for (int i=0;i

[@4444-c4s]

A little catch for Pattern-7. Don't need to print last spaces. Just 2 inner for loops will work.

[@ChetanRajput-ju2sd]

problem 22 - For Python: I found below approach most optimised. If u observe (row wise): 1st half and 2nd half is a mirror (ignore middle for a while.) 1st half of a row is a combination of two parts: a - fix part b - decremental part for eg , for n = 6 (considering just 1st half of a row) row1 : 666666 -> all fix , no decremental row2 : 655555 -> 6- decremental , 55555 - fix row3 : 654444 -> 65 - dec , 4444 - fix row4 : 654333 -> 654 - dec , 333 - fix derive the code for dec and fix , then combine the string to generate 1st part of a row , then neglet last character and reverse the result , concatenate both original and reverse to get full row. time difference (n = 500 ) 1. n - min() approach : 256ms 2. above approach : 54 ms

[@cinime]

Understood! Super amazing explanation as always, thank you very much!!

[@frekkytanish]

for pattern 16 : int n; cin>>n; for(char ch = 'A' ; ch < 'A'+ n ; ch++ ){ for(int cha = 'A' ; cha

[@user-jy3sc2df7y]

Is anyone doing it using Java?

[@xihadityajaiswal8797]

Me

[@harikrishnapatel7514]

yes

[@shaunakmajumdar5278]

Ofc

[@vikrant3640]

Ofc😊

[@ProgramerRaj]

Yeah bro

[@yuuji-kun3061]

1:01:25 In Pattern18, we should not hardcode the value 'E', It should be dynamic. void pattern_18(int n) { for (int i=0; i

[@mx5enthusiast]

Exactly what I was thinking.

[@rishav144]

one of the best videos 🔥

[@GauravThinks]

bhai poori to dekh le pehle

[@yikes3807]

@@GauravThinks xd

[@itz-aksh]

pattern 19: we can also combine both upper and inner part by declaring variables for conditions and so avoid writing loops for 3 more times.. void print19(int n){ for(int i=1;in/2){ conditionstar=i-n/2; conditionspace=2*(n-i); } else{conditionstar=n/2-i+1; conditionspace=2*i-2;} for(int stars1=0;stars1

[@tgxgaming5880]

31:00 What if we don't print the last space ? Like, After the stars. Does it improve TC(Time Complexity)?

[@stith_pragya]

UNDERSTOOD......Thank You So Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@SoumeshRanger111]

Thank you for the video, If possible could you please share the tasks.json file here, i.e. how you're writing to the output file simultaneously.😊😊

[@neotetra0288]

thank you for this video, i think i need a lot of practice with patterns but this was good to get a basic idea.

[@Yash_Raj_J]

If possible can you tell us another method for pattern 22? I did not understand. BTW THANKS A LOT BHAIYA

[@BhuvaneswariSura]

import java.util.*; class Pattern { public static void main(String args[]) { int n=7; int c=((n+1)/2); for(int i=0;i

[@lavanya_m01]

Understood! Can't thank you enough, God bless you for your wonderful contribution to the community

[@Sagardeep_Das]

For pattern 18 this can be tried too. It can print decreasing pattern for any no of alphabets (don't exceed n = 25 in input) void pattern18(int n) { int count = n; for(int i = 0; i

[@yogeshverma7366]

Hey Striver, for the pattern 18, I guess the code will not work for n > 5, it will start printing characters lesser than 'A'.

[@varunsde9533]

to make it as per any value of n, make the following changes: char ch = (char)((int)'A'+n-i-1); for(int j=0;j

[@amansinghrajput18]

Understood..Great way of teaching

[@KeepCoding69]

Better solution for Pattern no.12 (using only 2 for loop) : #include using namespace std; int main() { int n = 9; int start = 1; int end = n*2; int temp; for (int row = 1; row

[@Manishgupta200]

The last one patttern is so tricky. Thanks, for the complete tutorial.. Very optimized way you explained to solve pattern of any type in a very easy way. Thankyou Striver.

[@nihalwaseem7106]

Started this playlist on 22nd sep 2024, great video covering various types of patterns questions

[@aayushjain3731]

Pattern 11 can also be written using a matrix like concept under the inner loop if((i+j)%2 == 0) cout

[@mintobit]

As an alternative solution for pattern #22: every number equals to the distance from center + 1. private static void numberPattern(int n) { for (int i = -n+1; i < n; i++) { for (int j = -n+1; j < n; j++) { System.out.print(1 + Math.max(Math.abs(i), Math.abs(j))); } System.out.print(" "); } }

[@niranjanniru4115]

BRO GUD TUTOR FOR DSA AND AS WELL AS FOR C++..... WE NEED MORE UPDATEZ LIKE THIX AND THANKS FOR YOUR LECTURE BRO

[@graviton001]

Pattern 5 my way of doing :- void print5(int n){ int a,b; for(a=1; a=a; b--){ cout

[@sarthakpatwari7988]

The logic for last one is something out of the box, great work man!!

[@programmingwallah2905]

One of the best man in coding community

[@ravikiranm-en4jr]

These patterns have helped me a lot to understand the loops. Thank you!!

[@SmailOujaoura-ty8ui]

Nicely presented lecture and crystal clear explanations.

[@Codeeater-vq2pz]

Patern 5 alternate approach void seeding(int n) { for(int i=1; i= i ; j- -){ cout

[@raunakkumar2555]

believe me I watched the petterns videos earlier from another youtubers too but I myself wasn't able to solve a single pattern but this time this person made me do the complex patterns by my self.He is something different .Hatss off to youu bhaiya

[@PriyankaKumari-cm3kq]

Problem no. 18 code- void alphaTriangle(int n) { for(int i=0;i

[@ninadkulkarni4951]

for the pattern 22 we can also have another logic we consider each number as a boundary of a square starting from 4 as the outer boundary moving in to 1 and put 4 conditionals just like pattern 21 if(i==0 || i==2n-1 || if j==0 || j==2n-1) printf("4"); if(i==1 || i==2n-2 || if j==2 || j==2n-2) printf("3"); if(i==2 || i==2n-3 || if j==3 || j==2n-3) printf("2"); if(i==3 || i==2n-4 || if j==4 || j==2n-3) printf("2");

[@pramodkumarverma8475]

We have to make it for n You can't fix it to 4 only

[@pramodkumarverma8475]

With this method n conditional statements will be needed

[@ninadkulkarni4951]

@@pramodkumarverma8475 ya my bad Time bht le rahi hai ye approach Not at all optimised

[@bhaveshpatil9271]

Understood. Thanks!

[@kumaraditya5513]

A different yet similar approach to Pattern 11 (java) 39:22 int num = 1; for (int i=1; i

[@Mansi_Javascript]

one of the best video ever on patterns!

[@sunny-teaches-sometimes]

For pattern 18, instead of assuming output always begins with char 'E', we could do this instead. Character A is 65 in ASCII, so the ending character at start would be => 65 + n -1 Since we are then decrementing the ending character at the start of every row => 65 + n - 1 - i While printing characters on a row's columns, we are incrementing the character to be printed => ord(rowStart) retrieves ASCII for the starting character for that row. ord(rowStart)+j retrieves the ASCII for the next character to be printed in that row chr(ord(rowStart)+j) retrieves the character for the respective ASCII Here's the python code for this pattern def pattern18(n): for i in range(n): rowStart = chr(65 + n - 1 - i) for j in range(i + 1): print(chr(ord(rowStart) + j), end='') print()

[@NerdOverPhysics]

for pattern 18 it will always start from 'E'. Instead, we can code this way better : void pattern18(int n){ char ch = 'A'+n; for(int i=0;i

[@nikhilmalhotra1986]

really glad a time you spent to explain this small topic in depth 🙂

[@susmitha5g627]

Really happy to get it all for free! Thank you so much Striver!

[@_TashfiqTaif]

IN problem 7 : No i think no need last space . I mean, space and star print the same thing .. Thank you sir code: #include using namespace std; #define ll long long int main() { int n; cin>>n; for(int i=1;i

[@krishnagarg2583]

for the 5th pattern you can also do for (int i = 1; i = i; j--) { cout

[@ankushshandilya]

Awesome Sir! Its truly insightful and interesting.