Lexicographical Numbers | Simple DFS | Leetcode 386 | codestorywithMIK

[@AmandeepSingh-uq3wp]

I also wrote the same code but still comes here to expand my knowledge from mik sir.

[@Playvish]

watched the video till 6:40 and solved the question successfully

[@beenasatyendrapal6405]

exactly....

[@AdarshSingh-cd8qt]

Bhaii bht accha explain kiya aapne thanks

[@ugcwithaddi]

Already done. Here to support 😊

[@MohammedArmaanUddin]

Same

[@peterfromengland8663]

Haddi ur leetcode I'd??

[@gui-codes]

aaj wala easy tha bhaiya. kuch mistakes karne k baad I was able to solve.

[@rockykumarverma980]

Thank you so much Bhaiya ji 🙏🙏🙏

[@best4567]

Another approach (using comparator) class Solution { public: vector lexicalOrder(int n) { auto lambda = [&](int &a , int &b) { string n1 = to_string(a); string n2 = to_string(b); return n2 > n1 ; }; vector nums; for(int i=1 ; i

[@ckshenoy]

Even I did the same way but this takes nlogn TC. The approach suggested in the video is actually better

[@bunnypubg3475]

Sorting nlogn lega

[@aliakbaransaria3-925]

very good explanation

[@harshitgupta6554]

Another approach using comparator function bool cmp(int a,int b){ string s1 = to_string(a); string s2 = to_string(b); int i = 0; while(s1[i]==s2[i]){ i++; } return s1[i]

[@bec_Divyansh]

I did this as well

[@rockykumarverma980]

thanks for sharing this approach🙏

[@harshitgupta6554]

@@rockykumarverma980 😇🙌

[@Its_Shubham_Negi]

This would lead to TLE since string comparision would take O(max(a,b)) for worst case

[@harshitgupta6554]

@@Its_Shubham_Negi no u can copy this code it's running fine and string length would be 5 at max as limit of n is 10^4 so it will take O(5) in worst case eventually O(1)

[@rev_krakken70]

I tried 4 approaches 1. First used basic sorting based on string comparison 2. Next approach I used tries to implement dfs 3. 3rd one I found out that trie is not required and used bfs 4. 4th one I used iteratively.

[@sourabhbindal774]

can you share the code of 3rd and 4th approach

[@salmaniproductions1104]

Thank you bhaiyya

[@rishithp3077]

thank you!!

[@vikramramkumar2087]

Hey! Could you make a playlist for Fenwick Trees??

[@abhinay.k]

thanks

[@gui-codes]

Bhaiya please post today's POTD - Leetcode - 440 - K-th Smallest in Lexicographical Order

[@shivanshplays9218]

Hello MIK sir, here is the approach that I came up with, I think that this also takes the same time and space complexity, although could you please confirm? TC:O(n) SC:O(log10(n)) class Solution { public: void recur(int no, int n, vector& arr) { if (no > n) return; for (int i = no; i < no + 10 && i

[@dayashankarlakhotia4943]

public int lexicalOrder(int n){ Listans=new ArrayList(); int cur=1; while(ans.size()

[@AlexKumarI-c6d]

bhaiya make a video on meet in the middle algorithm

[@gui-codes]

ye kya hai bhai ?

[@AkOp-bf9vm]

I TRIED SOMETHING WEIRD INITIALLY 😂😂😂😂😂😂😂😂. THIS IS THE EXAMPLE HOW NOT TO WRITE THE CODE USING CUSTOM COMPARATOR class Solution { public: vector lexicalOrder(int n) { vector result; for(int i=1;i

[@utube4026]

I thought the same😂😂

[@dayashankarlakhotia4943]

Ist🎉❤

[@jigarthakor3939]

what about iterative?

[@abhishek_0513]

Aaj wali video kis playlist mei hai ek playlist concept of recursion krke apne daali hai usme 18 videos hai usse alag bhi koi recursion playlist hai apke channel ki?

[@codestorywithMIK]

Here is the list - I usually create two playlists for every topic : 1) Concepts Playlist - Contains from basic concepts to expert concepts. 2) Popular Interview Problems playlist. I have created concepts and interview problems playlist for 1) Graph 2) Recursion 3) DP 4) Segment Tree Planning soon to create concepts playlist for Tree as well. Graph Concepts - kzbin.info/aero/PLpIkg8OmuX-LZB9jYzbbZchk277H5CbdY&si=lZG2IJTmSW4kRrx- Graph Popular Interview Problems - kzbin.info/aero/PLpIkg8OmuX-I_49pdy1XFY6OcATnxUrrO&si=CG2JvGWVmvoSqvWA Recursion Concepts - kzbin.info/aero/PLpIkg8OmuX-IBcXsfITH5ql0Lqci1MYPM&si=614iI4NyHY-FTeJH Recursion Problems (In progress) - kzbin.info/aero/PLpIkg8OmuX-IXOgDP_YYiJFqfCFKFBDkO&si=88fBhRnr62OYTnDP DP Concepts (In progress) - kzbin.info/aero/PLpIkg8OmuX-JhFpkhgrAwZRtukO0SkwAt&si=laFVYy6ep2BkOg0s DP Popular interview problems - kzbin.info/aero/PLpIkg8OmuX-L_QqcKB5abYynQbonaNcq3&si=VHEn9b-wqTnAVyyi Segment Tree Concepts - kzbin.info/aero/PLpIkg8OmuX-K1qUIQToCllUO0UIKXt8dB&si=xm7DqRN4H0eZwna4

[@abhishek_0513]

@@codestorywithMIK thank you so much bhaiya

[@manasdeora4601]

tried implementing same solution in java, gives TLE. public class Solution { public List lexicalOrder(int n) { List ans = new ArrayList(); for (int i = 1; i n) { return; } if(ans.contains(i)) { return; } ans.add(i); for (int j = 0; j n){ return; } solve(x, n, ans); } } }

[@captain-ne8qy]

if(ans.contains(i)) { return; } ❌❌ why this extra check? it takes O(ans.length()) each time ***************************************** for (int i = 1; i

[@yashwardhansingh8401]

sir how can we done this with O(1) space

[@gui-codes]

space agar dekha jae to O(1) hi hai. Editorial me dekho leetcode k - Given that the maximum value for n is 50,000, the maximum number of digits d is 5. Thus, the recursion stack depth and corresponding space complexity is O(d), which simplifies to O(log 10 ​ (n)), but with a maximum constant value of 5 for practical constraints. It can also be argued as O(1).

[@princeberi4501]

same code in java giving TLE.

[@kancharlasrimannarayana7068]

can we solve it in O(1) space conplexity?

[@gui-codes]

space agar dekha jae to O(1) hi hai. Editorial me dekho leetcode k - Given that the maximum value for n is 50,000, the maximum number of digits d is 5. Thus, the recursion stack depth and corresponding space complexity is O(d), which simplifies to O(log 10 ​ (n)), but with a maximum constant value of 5 for practical constraints. It can also be argued as O(1).

[@kancharlasrimannarayana7068]

thank you but if the constraint changes means? and if we change the question like this if we given a random array of numbers as input and asked to sort lexxigraphicak means how to do it?

[@brokegod5871]

@@kancharlasrimannarayana7068 I don't quite understand what you mean but if you are saying the array has random elements and you want to sort in lexicographical order, that is nothing but the usual ascending/increasing order sorting. Lexicographically usually means according to dictionary, so yeah 1 then 2 then 3 then 4 and so on. That is just normal sorting

[@kancharlasrimannarayana7068]

​@@brokegod5871, [1,10,3,5,22] as input normal sort output is [1,3,5,10,22] but lexicographical sort output is [1,10,22,3,5]

[@gui-codes]

@@brokegod5871 right

[@sujayvikramgs8588]

bhai aaj ka problem meh tho same method se solve liya tho TLE aagya time efficient nhi haina then why this T_T;

[@codestorywithMIK]

Yes if you use array to store it will give - MLE and if you solve without using array, it will give TLE because constraint is huge. We need logarithmic approach. The video is being uploaded now ❤️🙏

[@sujayvikramgs8588]

@@codestorywithMIK thanks bhaiya will watch the video now. thanks for all your efforts bhaiya :)

[@dibbodas4116]

didnt understand the space complexity

[@codestorywithMIK]

For recursion problems, we always check what is the maximum depth of the recursion tree. If you notice , for any recursion call, you won’t go more than digits deep in recursion. Please also refer to my Recursion Concepts playlist where I have explained how to calculate space complexity of recursion calls

[@dibbodas4116]

@@codestorywithMIK bhaiya i understand the recursion tree but why it is log base 10?

[@codestorywithMIK]

Because for any decimal number , the number of digits can be calculated by log(number) to the base 10. Let’s understand from an example : number = 234 To find the number of digits, keep dividing it by 10z 234/10 =23 23/10 =2 2/10 =0 If you notice how many times you had to divide 243 by 10 ? 3 Dividing by 10 until the number is 0 i.e. log10(number)

[@dibbodas4116]

@@codestorywithMIK thanks bhaiya now i understood

[@aizad786iqbal]

brute force hi kar diya tha, extra space use karke :D and sorting pe, somehow it passed... class Solution { public List lexicalOrder(int n) { List ans = new ArrayList(); List list = new ArrayList(); for(int i=1; i

[@prathmeshkakde3731]

from where did you learn dsa?