Sequential Digits | 2 Approaches | Leetcode 1291

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This is the 81st Video of our Playlist "Array 1D-2D : Popular Interview Problems".
In this video we will try to solve a very good Array Enumeration problem - Sequential Digits | Leetcode 1291
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Sequential Digits | Leetcode 1291
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Approach-1 Summary : The first approach utilizes a queue to generate sequential digits in a breadth-first manner. It starts by enqueueing digits from 1 to 8 and then iteratively dequeues a digit, checks if it falls within the given range, and adds it to the result vector. Additionally, it enqueues the next sequential digit formed by adding one to the last digit of the current dequeued number. This process continues until the queue is empty, and the result vector contains all sequential digits within the specified range.
Approach-2 Summary : The second approach uses a precomputed list of all possible sequential digits up to nine digits in length. It then iterates through this list, filtering out numbers that are below the given low value and above the high value. The result vector is populated with the filtered numbers.
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✨ Timelines✨
00:00 - Introduction
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[@ramanthakur2482]

my day doesn't complete before watching your POTD video....ur way of teaching is quite amazing

[@AnjuGupta-sz1uk]

Both approaches are very clear

[@souravjoshi2293]

" bhai bhai approach" was awesome 🤣

[@DevOpskagyaan]

Bhai travel k waqt video upload. Salute to your dedication 🔥❤️

[@aryanraj9235]

last one was funny 😂😂,Are bhai bhai bhai...

[@karanthakur11]

method 1: slight change (change from recur memo to one based indexing) ese bhi likh skte hein recusrion memoisation ko method 2:: ise pehle tabulation slight change from tabulation to space optimization 33:44

[@codeandtalk6]

❤❤❤❤

[@Mercer_077]

13:23 First approach bhi toh O(1) hi time le rhi hai , n cant exceed 36 for given constraints.

[@neerajchopra2758]

sir please make gfg podt today question implement Atoi explanation 😭

[@shreeramsingh7652]

sir, we need one detail video on while seeing any problem Constraints on any coding paltform by seeing the Constraints we should able to think which algorithm can apply and what its time complexity, Because i am seeing that in my class top coder are first he see question statement and directly they are seen Constraints and predict which algorithm i should apply ,they know basic cp, please kindly help us sir🙏🙏..

[@codestorywithMIK]

Definitely. Noted on this.

[@naveenkumarkota2090]

keep making this videos will learn from u and we will be support mike

[@AnkitSingh-tm5dp]

Best Explaination , Problem is also good i never do this type of problem of I have no idea of Intuition other then recursion.

[@girdharipatel2]

are bhai bhai approach 😂

[@codingdev4196]

Hacker bhai 😂

[@LakhanRathod1242]

Are bhai bhai bhai approach 🤣🤣🤣🤣

[@bhuppidhamii]

waiting for today's potd

[@codestorywithMIK]

kzbin.info/www/bejne/gqfZepaQjNmopbssi=HUQTNEfDaAWQaqsc ❤️❤️😇😇

[@Trendspra]

How can anyone explain programming logic so easily.....Hats off

[@voidPointer040]

Interviewer be like arayy bhai bhai bhai😂😂

[@souravjoshi2293]

[@DreamPuller-cf6yl]

need guidance video for your videos

[@amitmehraa]

man you are awesome😂😂

[@thor1626]

A sliding window approach which beats 100% users in java : class Solution { public List sequentialDigits(int low, int high) { String nums = "123456789"; int lowerRange = 0, upperRange = 0; int lowerTemp = low, higherTemp = high; while (lowerTemp > 0) { lowerTemp /= 10; lowerRange++; } while (higherTemp > 0) { higherTemp /= 10; upperRange++; } List seqDigits = new ArrayList(); for (int i = lowerRange; i = low && temp

[@Incredible2India]

2nd approch mst h 😂

[@gauravbanerjee2898]

thanks a lot bhaiya ❣❣ Hats off to your dedication🫡, providing videos even when you are travelling ❤❤

[@dipakdhakate5613]

bhai ka samzana like "so beautiful so elegant just looking like a wow"😅

[@ankurtiwari9621]

I just waited for your video after trying..Thanks for your efforts becoming a big fan ..☺

[@theOmKumar]

i solved it like this: brute force using sort vector sequentialDigits(int low, int high) { vector ans; for(int i = 1; i = low && res

[@wearevacationuncoverers]

waiting for you. ❣ and hats off to your dedication.

[@thefintechguy899]

"are bhai bhai bhai"

[@lofireverbz-wy7go]

bhaiya placement season aara hai june mai so please kuch new cheez le aoooo tadkta bhadakta like oops or something of your choice:)!!!!!

[@Idk-qg7hb]

I did bhai bhai approach 😂

[@souravjoshi2293]

same 🤣

[@sachinnarang2453]

Beats 100% class Solution { public List sequentialDigits(int low2, int high2) { int count = 0; int low=low2,high=high2; while(low>0) { low/=10; count++; } int firstCount = count; count=0; while(high>0){ high/=10; count++; } int secondCount=count; List lis = new ArrayList(); for(int i=firstCount;i

[@manimanohar_001]

Bro LC 1247 pe ek video pls🙏🙏🙏

[@S3aw0lf]

class Solution { public List sequentialDigits(int low, int high) { List result = new ArrayList(); for (int i = 1; i

[@sauravchandra10]

My approach: class Solution { public: long create(int start, int size){ long res=start; for(int i=start+1;size>0 && i=low){ ans.push_back(curNum); } } curSize+=1; } return ans; } };

[@thekindspill]

dedication level 🔥🔥🔥🔥

[@aws_handles]

Are bhai bhai bhai 😅👌🏻

[@naveenkumarkota2090]

like mardo sab log

[@PiyushSingh-xg4mc]

👌👌👍👍

[@Shakib.Siddiqui]

Bhai Bhai approach......is useful only for online assessment 😂

[@MohammadTouseeb]

Bhai ek hi dil hai kitne bar jeetoge. Ab agar koi question apne nhi kiya hota youtube pe Mai kisi or ka dekhta hi nhi chor deta hu bad me karenge😂

[@harshtiwari416]

bhai gfg ke count digits of a number ki bottom up video kab aayegi???

[@DevOpskagyaan]

I also am waiting . I think he is travelling and will upload when he is back. @codestorywithMIK - waiting

[@ugcwithaddi]

🔥🔥🔥

[@NikhilSatyam]

bhaiya pls try to upload this notes

[@priyanshugouravsarangi8553]

BFS FROM NOTHING HINT OF BFS FAIRS I THOUGHT OF DP FIRST

[@KumarMuravath-YearBTechChemica]

341 😳

[@manimanohar_001]

First view

[@dhairyachauhan6622]

i solved using digit dp but afterwards realized that the question was asking a vector. So there was nothing to memoize 🥲😁. But atleast this method is unique but complex and i dont know about the complexity but i guess it will be slower since i cannot memoize. Took my code (4ms) DIGIT DP (TEMPLATE)SOLUTION class Solution { public: void solve(string &R,int digit,int result,int prevDigit,int leading,bool tight,set&ans,bool erase){ if(digit == 0){ if(erase){ ans.erase(result); return; } ans.insert(result); return; } int ub = tight?(R[R.length()-digit]-'0'):9; for(int i = 0;i

[@souravjoshi2293]

nice man

[@tutuimam3381]

❤❤❤

[@mohdzaid1158]

are bhai bhai bhai approach 🤣🤣🤣🤣