L18. K-th Permutation Sequence | Leetcode

[@takeUforward]

Please leave a small comment if you understand by spending your time here, every comment motivates me more :) C++ Code: github.com/striver79/SDESheet/blob/main/permutationSequenceCpp Java Code: github.com/striver79/SDESheet/blob/main/permutationSequenceJava Instagram(For live sessions): instagram.com/striver_79/

[@kajalKumari-yd9tj]

this code is not working for a test case on interview bit kzbin.info/door/z9dtkYt27fb2priOSu_3dw public class Solution { public ArrayList findPerm(int A, Long B) { Long fact = 1L; List numbers = new ArrayList(); ArrayList ans = new ArrayList(); for(int i = 1;i

[@shauryavatsa595]

It was unimaginable that i could ever learn DSA, but then i found you. Thanks a lot.

[@aakstatic]

This series is so underrated! It deserves a lot more views. Thanks to Striver for doing this noble work. :)

[@divyareddy7622]

why is it 16%6 at 9:23

[@hkedia08]

@@divyareddy7622 bcz he had divide them in group of 6 and he is taking 0 based indexing so 6+6+4 that is k th number hence 16/6=2 and 16%6=4 so (2*6+4)=16(kth number) .

[@Zunan263]

Actually 💯

[@asishcodes]

It's not underrated

[@lakshsinghania]

watched it 2 times to get the concept

[@anishali8042]

Please pin this comment! 8:52 Why 16%6? It is because there are 4 groups defined and in one of the 4 groups, the kth permutation will lie. 1, (2,3,4) Group 1

[@adityamaheshwari3046]

13:28 , 20:16 - Just a suggestion, as soon as we see the value of k is zero we can break/end recursion right there and append the remaining string into the answer. Intution is that k=0 means we have to find the first smallest permutation and first smallest permutation is the number itself (since our input string is already sorted)

[@atjhamit]

Works for the current scenario but not a good practice to introduce if statements for edge cases. Ideally the lesser the number of conditions, the better(code readability and other such standards).

[@lakshyaagarwal5301]

Have tried that on gfg and leetcode, it doesn't satisfy the time complexity. Think about the worst case scenario what if n=9 and k=9! i.e the required string is the last possible permutation It would take a lot of time. I know how it feels don't worry i have been there 🥲

[@YashHalgaonkar]

Much clear and conceptual code: string getPermutation(int n, int k) { int fact = 1; string res = ""; vector arr; for(int i = 1; i 0) { fact = fact/arr.size(); res += to_string(arr[k/fact]); arr.erase(arr.begin() + k/fact); k = k%fact; } return res; }

[@freecoder522]

can I get the complete code

[@YashHalgaonkar]

@@freecoder522 its complete.

[@freecoder522]

complete working code : #include using namespace std; string getPermutation(int n, int k) { int fact = 1; string res = ""; vector arr; for(int i = 1; i 0) { fact = fact/arr.size(); res += to_string(arr[k/fact]); arr.erase(arr.begin() + k/fact); k = k%fact; } return res; } int main() { int n = 3, k = 4; string kthsequence = getPermutation(n, k); cout

[@samratpodder545]

Thanks Striver, for sharing the pick / not-pick thought process and the clear demonstration of backtracking. This playlist is equally awesome for beginners as well as experienced DSAers. Recursion is now a lot clearer. Will be starting with the DP Playlist very soon !!

[@atharvakalele37]

This was a tough problem to grasp but u made us do anyway...u r awesome dude!

[@lakshsinghania]

indeed it was

[@ghj7275]

Completed the entire playlist and let me tell you, this is the best one out there. There are so many questions that I couldn't solve an year before but now I'm able to after watching this series❤❤❤

[@harshtewari4333]

Same

[@surajmisal3898]

I just have completed this series. It's awesome and I'm looking forward to complete graph series now.

[@Abhishek-do8mp]

graph series is awesome too..

[@saranshpathak3795]

Your logic are too good, the way you break problem into such smaller/easier part is amazing.

[@escape704]

For the same que in interviewbit take care of the overflow condition when calculating the faction above numver 11... here is the code--- long factorial(long n){ if(n==0 or n==1) return 1; if(n>12) return INT_MAX; else return n*factorial(n-1); } string Solution::getPermutation(int n, int k) { vector nums; for(int i=1;i

[@kira-nr3qj]

we can do this question using binary exponentiation to get a more optimal solution, code - vector applyPermutation(vector sequence, vector permutation) { vector newSequence(sequence.size()); for(int i = 0; i < sequence.size(); i++) { newSequence[i] = sequence[permutation[i]]; } return newSequence; } vector permute(vector sequence, vector permutation, long long b) { while (b > 0) { if (b & 1) { sequence = applyPermutation(sequence, permutation); } permutation = applyPermutation(permutation, permutation); b >>= 1; } return sequence; }

[@aakriti1]

Recursion series done🥳 Thanks a lot Striver Bhayiya! 😊

[@codingloop]

Completed entire playlist took 4-5 days and now I have good hold on Recursion & Backtracking . Thanks for your wonderful explanation and efforts

[@lakshsinghania]

same

[@CodeSuccessChronicle]

Clean, Clear, Well Structured Explanation Just Fantastic

[@divyareddy7622]

why is it 16%6 at 9:23

[@CodeSuccessChronicle]

@@divyareddy7622 it's been 4 months, I am unable to remember the pattern. I have to re-watch it again. But I did understand it so well, I'll definitely let you know after I watch it again and meanwhile I request you please watch again, this time you might get it. Try...

[@divyareddy7622]

@@CodeSuccessChronicle yess i did! He didn't explain properly the logic for 16mod6 ... I understood 16/6 but not 16mod6. Let me know when you're free!

[@madarauchiha-dr5ws]

@@divyareddy7622 here he did mod...reason is ... He initially divided the sequence in 4 groups. Now when he did / we know in which group our value resides. Now by doing % we now know at which place of that particular group this value resides. Hoping it is clear..let me know if not

[@anishali8042]

​@@divyareddy7622 8:52 Why 16%6? It is because there are 4 groups defined and in one of the 4 groups, the kth permutation will lie. 1, (2,3,4) Group 1

[@dsa_tutorial]

Thank You striver for this video actually i am generating all numbers and it is giving me TLE .Now I will Optimise the solution using this method .

[@SubashsVlogs]

Your content is the best one can get on KZbin 🙌

[@himadrinath1502]

Bhiya mai expert hu codeforces mai kuch guide dijia how to reach candidate master my maximum rating in cf is 1700+ and I also 6* rating in codechef via short contest and 3 to 4 long contest is there very high leve genius Iq needed to reach very good rating in codeforces like that 2600+ in codeforces or any extra smart work needed to achive this beacuse some people are very short time to reach 2400+ rating in codeforces

[@Aks-47]

@@himadrinath1502 its evident from your what your saying...expert and 6 star ,my foot lol

[@messyronaldo3161]

@@himadrinath1502Kuchh jyada hi kamjor nhi h tu🤕🤕

[@himadrinath1502]

Tumara baat samja nahi bhai keya bola????

[@himadrinath1502]

Keya bola bhai??

[@sauceopet]

this content is gold. but this man ........... diamond !

[@mandeepkaur749]

these series are really great. A lot helpful and are a savior

[@namankeshari7332]

Extremely nicely explained! From brute force to most optimised approach with this detailed explanation, absolutely loved it!!!

[@praveenrawat6574]

a very similar ques to this came in atlassian oa at my college, was able to solve this coz of u

[@LegitGamer2345]

i think this is the best explanation for this problem available on the internet EDIT : i think this is the best explanation for this problem available out there

[@takeUforward]

Glad it helped!

[@sanskarmaliwad4240]

Loved the frequency of videos.Thank you bhaiya

[@stith_pragya]

Understood,,,Thank you so much for this wonderful video....🙏🏻🙏🏻🙏🏻

[@Tomharry910]

Wonderful explaination. You walked us through the whole process with so much ease and clarity. Thank you for the good work!

[@sriharika2775]

Thank you so much for teaching in a simpler and understandable way..

[@parthsalat]

This one's complicated for me. I'm gonna keep this in my todo list.

[@PraddyumnShukla]

I think there's a mistake. In the second iteration when k = 4 and array = {1 , 2 , 4} then the 4th sequence will lie in the second block(the middle one) and not the in the last block. Because, the 3rd and 4th permutation is in the middle block whereas the 5th and 6th permutation is in the last block.

[@swetajha2265]

We have to find the 17th element. 12 elements are already covered in the first iteration. Now we have to find the 5th element that would lie in the last block. Since it is 0 based, this confusion is there.

[@luckydewangan5985]

Acutally he is considering 0 based indexing that's why you are consfused, by k=4 means 4th index ie 5th sequence

[@PradyumanSharma-s4c]

Bhai hamne jab k=17 ko 16 bnaya, tabhi hamara k 0-indexing ka hissa ho chuka tha, meaning now our k is 0-indexed already. Hence k=4 lies in the third permutation set {1,2}

[@pallavivarshney1744]

Thanks it was really helpful. I have been following your sheet and it is really good. Please add more DP problems covering all categories of dynamic programming.

[@nikhilrajsoni4541]

more formal way (taught by striver in his graph seiries): class Solution{ public: vectorx={1,0,0,-1}; vectory={0,-1,1,0}; string d="DLRU"; void solve(int i, int j,vector&a,int n,vector&ans,string move,vector&vis){ if(i==n-1&&j==n-1){ans.push_back(move);return;} for(int k=0;k

[@irrompible4588]

Amazing Explanation. I got placed because of learning how to explain recursion ..Your content has the biggest contribution..

[@gauravshukla6675]

which company??

[@irrompible4588]

@@gauravshukla6675 it's PeopleStrong

[@anirudhatalmale5575]

Only one giving solution to problems which are very important and not being attempted by anyone else on youtube

[@maneeshguptanalluru7807]

Another approach to this problem: 1. Sort the array 2. Use the next permutation code as the subroutine 3. Repeat it k times to get the kth permutation TC - O(k*n)

[@rishavsaha5254]

@@prashantthakur9329 no the time complexity for next_permutation is O(n). His answer is also correct, but it's naive as in interview we are not expected to use such kind of STL. Like, if in interview you are asked to find the count of set bits for a number. You definitely won't say about the __builtin_popcount() STL.

[@harshithravinoothala8157]

The value of k is 10^9 so it will be taking a lot of time.

[@amritanarang7305]

2 videos in one day👍 osm bhaiya

[@priyajitpaul4190]

This was a tough problem but thanks to Striver bhaiya,gave a beautiful explanation which made it understand easily.😀

[@vatalaaiswarya7956]

You are really taking us forward. Please do moe videos. Your explantion is the best. None of the other's videos helps me like yours did 😊

[@divyareddy7622]

why is it 16%6 at 9:23

[@-RohitJaiswal]

@@divyareddy7622 Because we need to find the 16th permutation and 16 can be written as floor(16/k)*k + 16%k, we have jumped the first part i.e. floor(16/k)*k no. of permutations by deciding the first place, now we are just left with jumping the 2nd part i.e 16%k permutations.

[@anishali8042]

​@@divyareddy7622 8:52 Why 16%6? It is because there are 4 groups defined and in one of the 4 groups, the kth permutation will lie. 1, (2,3,4) Group 1

[@subhajitdey135]

I think for the 3rd step, i.e. for {1,2}, k should be evaluated by 0%1 and not 0%0. Because the number of possible candidates is 1 each for {1,2}

[@satendra6200]

yes Correct it should be 0 % 1 since size of each is 1 not 0

[@lokeshnegi5051]

finished the entire series thanks striver for this wonderful session on recursion.

[@subhajit.kundu001]

man , how he catch the pattern is literally amazing

[@salmankhader1258]

The another approach could be you can think of next permutation problem which gives the next permutation in o(n) time so you can call that function simply k times. it will be done in o(n*k) time

[@sagardutta799]

2 reasons why using STL library is not promoted in this ques: 1. Whenever we write an algorithm for a code, it has to be universally applicable to all languages. Not all languages have in built next_permutation function 2. in worse case, the time Complexity will become O(n!) where n when 9 will make the program run for millions of times

[@arpitomar7875]

Great explanation learned a lot from this series.

[@akhileshsaga217]

Thanks Striver, for amazing playlist on recursion.

[@kanikachauhan-g2g]

How come with these excellent teaching skills? Awesome playlist for recursion 🔥

[@AshutoshAnand-o5l]

God bless this guy!!

[@bidishadas832]

This is the best explanation of this problem.

[@cricketwala937]

I love the way you explains problem 😍

[@harsimransingh8901]

Your work is really providing, Appereciable🙌

[@abhinavsrivastava2128]

Awesome explanation 👏. However the Java solution might not result in O(N) space if we dont keep ans as a StringBuilder. Not all languages are as bestowed as C++ where adding char to string is O(1)😅😅

[@abhisheksa6635]

Need to check way more times for this logic to set in.

[@boostedsilverplays]

Man I feel like a genius after understanding the math behind the solution damn! Thanks raj 🙌🏻

[@RaubinshArya]

Completed this entire series and after watching this video I am going to request Leetcode to please mark this question as Easy not Hard. 😀

[@RajnishPrasadKalwar]

Jab etna aasan tha to yaha q bhai khud karleta

[@nehasingh9795]

Thankyou so much for making the best recursion playlist

[@vasujain1970]

This is just gold

[@037_sayedramishali7]

Khud se try kara 2 hrs mai nhi hua , video dekh ke smaj aaya thanx bhai🙏🙏

[@tanuagrawal9921]

Your explanation is best !!

[@harshgarg7025]

Hey, we can optimize the time complexity to O(nlogn) by using other data structures instead of vectors or arrays.

[@vaishnavisood9699]

how?

[@rakshitpandey7517]

to store in sorted arrangement, it will be requiring NlogN to maintain the sorted order anyhow. You can't reduce that..

[@gauravsharma-ho5hx]

recursive soln:- int fect(int n){ int product=1; for(int i=1;i

[@indiansoftwareengineer4899]

Thanks bhai, hope you are well now... great knowledge for free...

[@nitishkumargupta2787]

Bruh..how do you explain so flawlessly

[@dhrroovv]

completed the playlist ... thanks

[@willturner3440]

You're doing great brother 🤗

[@debasishphukon5345]

Great Explanataion as always, I also wrote a recursive code for this, you guys can check it out : void findPermutation(string &ans, int k, int n, int st) //st is starting index { if(k==0) { return; } int f = factorial(n-1); int i = k/f; swap(ans[st], ans[st+i]); sort(ans.begin()+st+1,ans.end()); //O(nlogn) findPermutation(ans, k%f ,n-1,st+1); } I did swap instead of removing it from the array, this could be improved further by using erase function that striver used.

[@bhaveshkumar6842]

@2:47 does making a deep copy involve converting an array of integers to an integer? Because to generate the permutations for 123, we'll use {1, 2, 3} and swap the array elements, so is deep copying converting an array of integers to an integer?

[@ritikshandilya7075]

Thanks for wonderful session Striver 🐯

[@ghoshdipan]

Most awaited video on recursion!❤️

[@dhanuufcdk]

Awesome series. Really recommend it

[@utkarsh_108]

this ques somewhere pertains to Class 10th NCERT Maths book questions given in Examples section(maybe example 22 or 23 afaik) of chapter PNC the ques was like this, "If the letters of the word 'NAAGI' are arranged as in a dictionary then the rank of the given word...."

[@dhruvchoubey4713]

Do we have P&C in 10th? 😂

[@ushmitadutta6713]

Completed the entire series

[@spyrowolf2211]

to come up with the optimal solution in a 30 minute interview is not "very simple" Striver, but regardless thank you for the walk-through of the best approach

[@divyareddy7622]

why is it 16%6 at 9:23

[@anishali8042]

@@divyareddy7622 I hope you got the answer to why 16 % 6. Now please explain me that question's answer :(

[@yuvraj2038]

THANKYOUU SO SO MUCH FOR THIS SDE PLAYLIST :)

[@arnabchakraborty246]

Bhaiya Next Level ka explanation. Thank you sooo much

[@shubhamkumar3198]

Awesome playlist

[@shubhiagarwal4047]

yes this sheet is really underated please striver bhaiya complete your sde sheet as soon as possible i have seen almost all ..

[@devanshupadhyay2658]

the best playlist ever.

[@satnamsingh3801]

Superb explanation 😊

[@rajeevnathani1153]

Can we solve it using recursion like function calling the same function, without using the while loop ?

[@AyushSachan2211]

Yes

[@rohankumarshah5679]

int fact(int n) { if (n>12) return INT_MAX; int f = 1; for (auto i = 2; i fact(n)) return ""; int f = fact(n-1); int pos = k / f; k %= f; string ch = to_string(numlist[pos]); numlist.erase(numlist.begin() + pos); return ch + backtracking(k, numlist); } string Solution::getPermutation(int n, int k) { vector numlist; for (auto i = 1; i

[@iamnoob7593]

Amazing series.

[@Aman_Gupta54]

damnnnn that was a very good explanation dude. keep it up

[@mohammedwaseem8599]

Hello bhaiya i hope you are doing extremely well

[@prashanthreddy5006]

Great Work bro Helping me a lot

[@saranghae3720]

C++ Code class Solution { public: string getPermutation(int n, int k) { int fact = 1; vector nums; for(int i = 1; i

[@vsrproduction1875]

Thank you very much , it was helpful.

[@kms8320]

Great code👏👏👏

[@ayeshasumaiya8280]

Wonderful Explaination Bhaiya!!!!

[@arpanbanejee5143]

Excellent explanation! Don t know why there are so few likes!

[@rajmehta6498]

Thank you so much,great explanation!!

[@_hulk748]

Thankyou sir❤🙏🙇‍♂

[@nimitjain660]

wonderful solution ans so easy!

[@priyanshusain2274]

These type of questions require more memorization . So , we have to memorize the pattern of such type of questions?

[@raj_kundalia]

Thanks for this!

[@atharvakulkarni9033]

first leetcode hard problem that I was able to solve by myself

[@rachitagrawal3570]

Thanks for helping.

[@shubhamdeshmukh9287]

very good, awesome one

[@vaalarivan_p]

3:54 - 5:31

[@LegitGamer2345]

13:45 shouldnt it be 0%1 instead of 0%0

[@LegitGamer2345]

also , we can use hashmaps and doubly linked lists so that remove becomes O(1) , here we have distinct numbers so for every number we can use it as key in hashmap with value as node , so we can get the node at any moment , after getting the node we can achieve remove in O(1)

[@prasunkirtania4221]

Yes bro fact =1 at there .. so 0/1 = 0

[@vamsikrishnasai1682]

Can we take all permutations using recursion and sort the result array and find the k th permutation?

[@tushargogiya4017]

we can but its complexity is too High

[@tasneemayham974]

Hey, Striver!! I have a problem. Whenever I see your logic I can write the code easily. But the problem is it takes me a lot of time to figure out the logic and sometimes I don't know it. What do you think I should do to improve? May anyone help please. AND THANK YOU A MILLION FOR THE AMAZING SOLUTIONS AND A GREATER SERIESSS!!!!!!!!!!!!! 🔥🔥🔥👏👏

[@codeshwarkalpesh]

Keep it bro I love your videos

[@tanmaysinghal8370]

nice solution man, i'm thinking y u put this in the recurssion series?

[@shrad6611]

Java Version of Code: public class Solution { public String getPermutation(int n, int k) { List numbers = new ArrayList(); StringBuilder sb = new StringBuilder(); // create an array of factorial lookup int fact = 1; for(int i=1; i