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)
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
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.
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).
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; }
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; } }
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
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; } };
#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; }
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)
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
@@KartikeyTT it would work but using push-back gives better space complexity.
@@pranavmisra5870 bro space complexity does not depend on that
thanks
thanks brother
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.
Great sir, I just started downloading and you just uploading these videos which Faster than BSNL
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).
Exploring those Edge Cases and handling them is not everyone's cup of tea ❤🔥@Striver Kudoos!!
This question is king of edge cases
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;
}
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
Understood everything striver 😀😀
JAI HIND veere | darna nhi h | ek call kar kabhi bhi available 24/7 for you veere
I think time complexity will be O(4N)+O(K) .Inner while loop is also contributing o(N) time
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;
}
}
Mast maza agaya
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
thanks bhaiya
So many edge cases.........Uffffff
yeah, a lot of 'em
UNDERSTOOD;
Hey,striver we wants Heap playlist.please....
Understood
Thanks
tysm sir
The smallest number should be 1122 instead of 1219?
you cannot change the order
here order should be maintained, we cannot change the position
understood
Hi but how to think that we have to use stack to solve this problem🙂
you dont need to, you can do this by creating a new string and operating on it
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;
}
};
Memory Limit Exceeded
Can anyone please explain why he minus 0 in while condition during comparison of stack top with current character of string
To convert char to int.
It will work even if you don't do that
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;
}
};
Solution without stack:
class Solution {
public:
string removeKdigits(string num, int k) {
string newstr = "";
for(const char& c: num){
while(!newstr.empty() && c0){
newstr.pop_back();
k--;
}
newstr.push_back(c);
}
while(k>0){
newstr.pop_back();
k--;
}
int i = 0;
while(newstr[i] == '0'){
i++;
}
newstr = newstr.substr(i);
if(newstr == "")
return "0";
return newstr;
}
};
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";
}
}
your new lectures are becoming difficult to understand
but topics are also complex
this should have been easy if you've been following his playlist
Nobody asked for your opinion @@shreyxnsh.14
Nobody asked for ur opinion @@shreyxnsh.14
#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;
}
Understood
tysm sir
understood