@ManaswiRane
Day 33
@hypewaali
Already done. Here to support 😊
@rishithp3077
thank you!!
@rockykumarverma980
Thank you so much Bhaiya ji 🙏🙏🙏
@manasdeora4601
tried implementing same solution in java, gives TLE.
public class Solution {
public List<Integer> lexicalOrder(int n) {
List<Integer> ans = new ArrayList<>();
for (int i = 1; i <= n; i++) {
solve(i, n, ans);
}
return ans;
}
private static void solve(int i, int n, List<Integer> ans) {
if (i > n) {
return;
}
if(ans.contains(i)) {
return;
}
ans.add(i);
for (int j = 0; j <= 9; j++) {
int x = (i * 10) + j;
if(x>n){
return;
}
solve(x, n, ans);
}
}
}
@gui-codes
Bhaiya please post today's POTD - Leetcode - 440 - K-th Smallest in Lexicographical Order
@AdarshSingh-cd8qt
Bhaii bht accha explain kiya aapne thanks
@dibbodas4116
didnt understand the space complexity
@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<int>& arr) {
if (no > n)
return;
for (int i = no; i < no + 10 && i <= n; i++) {
if (i != 0) {
arr.push_back(i);
if (i * 10 <= n) {
recur(i * 10, n, arr);
}
}
}
return;
}
vector<int> lexicalOrder(int n) {
vector<int> arr;
recur(0, n, arr);
return arr;
}
};
ps: it took me 2 hours to come up with this approach :,)
@gui-codes
aaj wala easy tha bhaiya. kuch mistakes karne k baad I was able to solve.
@prathmeshkakde3731
from where did you learn dsa?
@dayashankarlakhotia4943
Ist🎉❤
@princeberi4501
same code in java giving TLE.
@kancharlasrimannarayana7068
can we solve it in O(1) space conplexity?
@AlexKumarI-c6d
bhaiya make a video on meet in the middle algorithm
@AmandeepSingh-uq3wp