Lexicographical Numbers | Leetcode 386

[@SamirKumarRakshit-c4z]

Perfectly expanded

[@techdose4u]

Thanks :)

[@AR_--007]

Really great explanation !

[@techdose4u]

Glad you liked it!

[@gauravsingh-mf9qf]

excellant ek bar dekh ke smjh aa gaya

[@techdose4u]

Great

[@niketandoddamani1093]

Perfect ! to the point thanks for the code walk through and the recursion tree

[@techdose4u]

welcome :)

[@JiyaRana-w7m]

Perfect i understand thank u sir

[@techdose4u]

Great

[@AnuragKumar-jo4li]

really good video

[@techdose4u]

Thanks

[@PreyumKumar]

depth will change acc. to number.

[@techdose4u]

depth will change in LogBase10 order. Since, problem mentions constraints to be very low hence we can ignore Call stack. Else, you can convert this logic to iterative approach :)

[@rishabh._.raj1]

Recursion will take stack space, this isn't an O(1) solution

[@techdose4u]

1. You can convert the logic to iterative code. 2. The depth is only max 5 hence can be taken constant.

[@rishabh._.raj1]

@@techdose4u True got it

[@nikhilaggarwal9325]

please provide java code as well.

[@Sehnaaz-vf7md]

The Java solution: class Solution { public List lexicalOrder(int n) { List ans= new ArrayList(); for(int i =1; in){ return; } ans.add(Current); //Generate the next number to add and check whether it exceeds the limit 'n' //remember the recursive tree while coding refer KZbin TechDose 386 for(int i=0;i n){ return ; } dfsHelper(next,n,ans); } } }

[@techdose4u]

Thanks :)