While pushing the st.top() to the res, use res.push_back(st.top()) instead of res = res + st.top() for leetcode, otherwise you'll get a memory limit exceeded error. If I have to guess, it's because string append (+) has a O(string size) where a new string is created and then appended whereas push_back is only O(1)
@KartikeyTT4 ай бұрын
my code worked fine even after using res + st.top() //here is the code class Solution { public: string removeKdigits(string nums, int k) { stack st; for(int i=0; i
@pranavmisra58703 ай бұрын
@@KartikeyTT it would work but using push-back gives better space complexity.
@KartikeyTT3 ай бұрын
@@pranavmisra5870 bro space complexity does not depend on that
@glyoxal19332 ай бұрын
thanks
@sukhii02202 ай бұрын
thanks brother
@206-tusharkantimukherjee327 күн бұрын
This question is king of edge cases
@manishnaidu303 ай бұрын
Hey striver, learning so much from this series, thank you first of all, and another issue I wanted to tell is that there are no articles and linked youtube videos in the website, Hope they are done soon as you are doing this series so that it is helpful for people who haven't checked this playlist on youtube yet.
@Roshan__30064 ай бұрын
Great sir, I just started downloading and you just uploading these videos which Faster than BSNL
@ShahNawaz-cx3pi3 ай бұрын
In some problems , brute force solution is more difficult that optimal solution in terms of implementation. BTW------ Brute force solution: To remove k digits from the number, we can explore all possible combinations and choose the smallest number. This involves generating all subsequences of length n - k and selecting the smallest one For Generating all the subsequences you can watch the striver's recursion playlist (There he has taught pick & non-pick method , after learning that method, recursion on arrays will become cakewalk).
@NetajiSaiSuru3 ай бұрын
Exploring those Edge Cases and handling them is not everyone's cup of tea ❤🔥@Striver Kudoos!!
@trashcan-md7cw2 ай бұрын
Instead of using a stack of characters, It is better to use a String as a stack like we used a list as a stack in asteroid collision... So we need not reverse the stack just remove the leading zeros and return it :) Happy Coding:) CODE string removeKdigits(string num, int k) { string st = ""; // Use a string as the stack int n = num.length(); // Traverse each digit in the string for (int i = 0; i < n; ++i) { // Pop digits from the stack if they are greater than the current digit // and we still have digits to remove (k > 0) while (!st.empty() && st.back() > num[i] && k > 0) { st.pop_back(); k--; } // Push the current digit to the stack st.push_back(num[i]); } // If k digits were not removed, remove from the end of the stack while (k > 0) { st.pop_back(); k--; } // Remove leading zeros int start = 0; while (start < st.size() && st[start] == '0') { start++; } // Get the resulting number without leading zeros string result = st.substr(start); // If the result is empty, return "0" return result.empty() ? "0" : result; }
@darkwarrior67672 ай бұрын
Instead of using stack to store answer you can directly use string and do operations on it using push_back and pop_back and back
@amitpandey83823 ай бұрын
I think time complexity will be O(4N)+O(K) .Inner while loop is also contributing o(N) time
@apmotivationakashparmar7222 ай бұрын
Understood everything striver 😀😀
@radhepatel68762 ай бұрын
Java Code: BTW This question is hard not medium class Solution { public String removeKdigits(String num, int k) { Stack st = new Stack(); String result = ""; for(int i=0;i0 && (num.charAt(i)-'0')0){ st.pop(); k-=1; } if(st.isEmpty()){ return "0"; } while(!st.isEmpty()){ result+=st.pop(); } String res = ""; int index; for(index=result.length()-1;index>0;index--){ if(result.charAt(index)!='0'){ break; } } for(int i=index;i>=0;i--){ res+=result.charAt(i); } return res; } }
@PawanKumar-hq6dy4 күн бұрын
last reversal we can avoid if we use deque where we can add and remove from first and last
@TOI-700Ай бұрын
JAI HIND veere | darna nhi h | ek call kar kabhi bhi available 24/7 for you veere
@shamanthhegde28202 ай бұрын
Hey Striver, I was able to solve this on my own thank you for that. to make it more simpler i removed the leading zero while inserting the numbers... the logic is in case the stack is empty and i am going to insert a zero that doesn't make sense because that number is going to turn out to be leading zero that's all public String removeKdigits(String num, int k) { Stack st = new Stack(); int charPopped = 0; String ans = ""; for(char ch:num.toCharArray()) { while(!st.isEmpty() && st.peek() > ch && charPopped < k) { st.pop(); charPopped++; } if(st.isEmpty() && ch == '0') continue; st.push(ch); } while(!st.isEmpty() && charPopped < k) { st.pop(); charPopped++; } if(st.isEmpty()) return "0"; int n = st.size(); for(int i=0; i
@killerboy23874 ай бұрын
Hey,striver we wants Heap playlist.please....
@yoddha621Ай бұрын
Mast maza agaya
@saumay-z14 ай бұрын
So many edge cases.........Uffffff
@shreyxnsh.142 ай бұрын
yeah, a lot of 'em
@oyeesharmeАй бұрын
thanks bhaiya
@DeadPoolx1712Ай бұрын
UNDERSTOOD;
@tamoghnasaha26672 ай бұрын
The smallest number should be 1122 instead of 1219?
@ayushaggarwal906Ай бұрын
you cannot change the order
@manas4656Ай бұрын
here order should be maintained, we cannot change the position
@KartikeyTT4 ай бұрын
tysm sir
@SibiRanganathL3 ай бұрын
Understood
@subee1283 ай бұрын
Thanks
@Shivi325903 ай бұрын
understood
@BeAcoder10114 ай бұрын
@shreyxnsh.142 ай бұрын
Thanks, was able to do this by myself, here is the C++ code: class Solution { public: string removeKdigits(string num, int k) { stack st; for(const char& c: num){ while(!st.empty() && c0){ st.pop(); k--; } st.push(c); } while(k>0){ st.pop(); k--; } string res = ""; while(!st.empty()){ res.push_back(st.top()); st.pop(); } reverse(res.begin(), res.end()); int i = 0; while(res[i] == '0'){ i++; } res = res.substr(i); if(res == "") return "0"; return res; } };
@Messi234853 ай бұрын
Can anyone please explain why he minus 0 in while condition during comparison of stack top with current character of string
@shashank_08073 ай бұрын
To convert char to int.
@valendradangi18223 ай бұрын
It will work even if you don't do that
@charuprabha87143 ай бұрын
Hi but how to think that we have to use stack to solve this problem🙂
@shreyxnsh.142 ай бұрын
you dont need to, you can do this by creating a new string and operating on it
@arunraj4383Ай бұрын
@navinvenkat3404Ай бұрын
C++ sol => class Solution { public: string removeKdigits(string num, int k) { stack st; int n = num.size(); for(int i=0;i0 && (st.top() - '0') > (num[i] - '0')){ st.pop(); k--; } st.push(num[i]); } while(k>0 && !st.empty()){ st.pop(); k--; } string result = ""; while(!st.empty()){ result += st.top(); st.pop(); } reverse(result.begin() , result.end()); int start = 0; while(start < result.size() && result[start] == '0'){ start++; } result = result.substr(start); return result.empty() ? "0" : result; } };
Java Code class Solution { public String removeKdigits(String num, int k) { Stack st = new Stack(); for(int i=0;inum.charAt(i)-'0' && k>0){ st.pop(); k--; } st.push(num.charAt(i)-'0'); } while(k>0 && !st.isEmpty()) { st.pop(); k--; } String res=""; while(!st.isEmpty()){ res+=st.pop(); } StringBuilder sb = new StringBuilder(res).reverse(); while(sb.length()>0 && sb.charAt(0)=='0'){ sb.deleteCharAt(0); } return sb.length()>0?sb.toString():"0"; } }
@MJBZG3 ай бұрын
your new lectures are becoming difficult to understand
@satyen46593 ай бұрын
but topics are also complex
@shreyxnsh.142 ай бұрын
this should have been easy if you've been following his playlist
@sanatgupta15622 ай бұрын
Nobody asked for your opinion @@shreyxnsh.14
@sanatgupta15622 ай бұрын
Nobody asked for ur opinion @@shreyxnsh.14
@valendradangi18223 ай бұрын
#include using namespace std; class Solution { public: string removeKdigits(string num, int k) { // char imp stack st; int n = num.size(); int removed = 0; for (int i = 0; i < n; i++) { while (!st.empty() && removed < k && st.top() > num[i]) // Don't do >= Dry Run 1,2,2,2,2,5 // k > 0 && st.top() - '0' > num[i] > '0' { removed++; st.pop(); } st.push(num[i]); } while (removed < k) { st.pop(); removed++; } if(st.empty())return "0"; // Makes code faster anyways // second last if can handle it int idx = st.size(); string ans(idx, '%'); // Prevents reversing the string while (!st.empty()) { ans[--idx] = st.top(); st.pop(); } int i = 0; while (i < ans.size() && ans[i] == '0') i++; // The erase function removes i characters // from given index (0 here) ans.erase(0, i); // If are using reversal // while(ans.size() > 0 && ans.back() == '0') // ans.pop_back(); if (ans.empty()) return "0"; return ans; } }; // This is a good question to study edge cases. // 1) k char may not be removed // 2) ans may contain leading zeroes // 3) // TC => O(N) + O(K) + O(N) + O(N) // SC => O(N) + O(N) int main(){ return 0; }