Codechef | Starters 140 | Tree Removal | Break The String | Editorial | Simple Solution.

Codechef | Starters 140 | Tree Removal | Break The String | Editorial | Simple Solution.
Tree Removal Solution : www.codechef.com/viewsolution...
Break The String : www.codechef.com/viewsolution...
CODECHEF
STARTERS
140
STARTERS140
TREEREMOVAL
BREAKTHESTRING
Codechef
Starters 140
Tree Removal
Break The String
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00:00 Tree Removal
18:57 Break The String

[@ronakkriplani1838]

nice observation in Tree removal

[@pretestpassed157]

Thankyou

[@gyanikumari4973]

Was waiting for this video. Thanks for uploading

[@pretestpassed157]

Thankyou

[@ronakkriplani1838]

Ohh thanks apne dono upload kar diye🙏🙏

[@pretestpassed157]

Welcome

[@anshumaan1024]

brother issi contest ka "tricky or treat" bhi bta do
mere partial test case hi chale ;(

[@adityatiwari80]

What a explanation brother , Appreciable .

[@pretestpassed157]

Thankyou ❤️

[@reactcode9148]

Nice Content

[@pretestpassed157]

Thankyou

[@kiranvats9503]

Sir please help

[@kiranvats9503]

Sir please please help me with this for fifth question giving run time error
I do not know string hashing so i use 3 vectors that store prefix strting suffix string and the current string with length n/2. Giving. Run time error on 5 th hidden test case . Took more than 5 hour
#include
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
string s;
cin>>s;
int ans=0;
int n=s.length();
if(n%2!=0) ans=0;
else{
int x=n/2;
vectorprefix(n),suffix(n),mera(n);
for(int i=0;i=0)
prefix[i]=prefix[i-1]+s[i];
else prefix[i]=s[0];
}
for(int i=n-1;i>=0;i--){
if(i

[@pretestpassed157]

1. What will be the overall size of the prefix or suffix ?
2. 1+2+3+4+.........+10^5 that is more then 10^10.
3. But we can't use the space complexity inside the function more than 10^6 and outside the function as more than 10^8.
4. So it's giving runtime error.

[@kiranvats9503]

​@@pretestpassed157 ok understood thank you very much sir.

[@kiranvats9503]

@@pretestpassed157
why is this giving tle and what is the expected time complexity for this question. Now i am not using the extra memory for prefix and suffix array
#include
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
string s;
cin>>s;
int ans=0;
int n=s.length();
if(n%2!=0) ans=0;
else{
int x=n/2;
string prefix="",suffix="",mera="";
for(int i=0;i

[@pretestpassed157]

What will be the size of mera and suffix. It's n/2
And for each iteration from 0 to n,
You are just checking either ( prefix + mera + suffix == s ) so this check itself will take n time complexity.
And this check is done inside each iteration from 0 to n so the overall time complexity will be n*n that is n^2 so it's giving TLE.

[@kiranvats9503]

@@pretestpassed157 ok got it. Is the time complexity of erase function o(1) here.