WSITES01 - Editorial

editorial
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21

Thanks @vijju123

22

https://www.codechef.com/viewsolution/13976703

Can any one tell me any tc on which this soln fails???

I have tried alot to solve this problem

23

I implemented it using nested dictionaries. I realise this is inefficient, but can someone tell me what testcase mine isnā€™t working on ?

https://www.codechef.com/viewsolution/13598055

24

Hi
I am using Trie for solving thisā€¦ Total 3 TCs are failing . Could somebody help meā€¦
https://www.codechef.com/viewsolution/14361002
Please help meā€¦

25

Hi I have tried all below TCs all are running fine . but I am getting WA in two TCs Please suggest any TC which I am missing

4

+ youtube
+ abcd
+ abcdchef
+ zzyy

Output
0

//// all negative

5
- abcdchef
- youtube
- youtubeisbest
- google
- googled
Output :
3
a
g
y

/// show strings in sorted order

6
+ google
+ googld
- googles
- googlea
- googlds
- googldr
OUTPUT :
4
googldr
googlds
googlea
googles

/////////
Any mathched -ve string shorter then positive

4
+ google
+ googld
- goog
- googles
OUTPUT :
-1

//////
Any mathched +ve string Longer then negative

3
- goog
- googles
+ google
OUTPUT :
-1

//same string its not allowed but still i checked this

2
- abc
+ abc
OUTPUT
-1
enter code here
26

this weird , the setterā€™s and testerā€™s solution link is just the link to editorial again

27

Yes. This editorial only has editorialistā€™s solution at the moment.

28

Hiļ¼Œanyone can help me figure out the problem below ?

the only difference for this two solution is

int n;
char c;
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
    getchar();
    c = getchar();
    cin >> s;

and

int n;
char c;
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
    cin >> c >> s;

the first one gets wrong answer, but the second one gets accepted.

the first solution link text

the second solution link text

but in local enviroment, I tested the first one , it can input and output normally with the samples.

thanks XD

29

@vicennial Why did u choose

int p=31; for calculating

ans=(ans*p+(i-ā€˜aā€™+2))%MOD;